{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0001","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eef767a26da2285b4f3c8bc429dd1bde2dd18ac8f3103977f98687772dce2da6","family_id":"qf-app21-orbit-fixed-audit","seed":202610213001,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":-7,\"b\":-9,\"modulus\":5,\"x\":0}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=-7, b=-9. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=0. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(5, -7, -9, 0, 1, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, -7, -9, 0, 1, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0002","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"23554f262901efe8d8cb6e24b4d37e22bf070f94815fe96e99c4bc144eb5cbfb","family_id":"qf-app21-orbit-fixed-audit","seed":202610213002,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":-6,\"b\":-7,\"modulus\":6,\"x\":1}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=-6, b=-7. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(6, -6, -7, 1, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, -6, -7, 1, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0003","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24f6de1b1970f1bd9d2e33df80a934a6976c6890e68ec7476ba78147d3da444e","family_id":"qf-app21-orbit-fixed-audit","seed":202610213003,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":-5,\"b\":-5,\"modulus\":7,\"x\":2}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-5, b=-5. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=2. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(7, -5, -5, 2, 6, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -5, -5, 2, 6, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0004","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7c3b301cd28786978b9fe71e01e1ba0b0c62ad6c3b0c0f0b3fedff3a2b540c8","family_id":"qf-app21-orbit-fixed-audit","seed":202610213004,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":-4,\"b\":-3,\"modulus\":8,\"x\":3}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-4, b=-3. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(8, -4, -3, 3, 1, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -4, -3, 3, 1, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0005","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f31dcd0f965bcfabc4976c600b7c97c6564c5f9e88c191b67552ed2fa596bea4","family_id":"qf-app21-orbit-fixed-audit","seed":202610213005,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":-3,\"b\":-1,\"modulus\":9,\"x\":4}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=-3, b=-1. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=4. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(9, -3, -1, 4, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, -3, -1, 4, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0006","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9646ba35befb7433d93c0ea3379b90c770f8f46b1828f077a0927de1280b61f1","family_id":"qf-app21-orbit-fixed-audit","seed":202610213006,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":-2,\"b\":1,\"modulus\":10,\"x\":5}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=-2, b=1. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=5. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(10, -2, 1, 5, 1, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, -2, 1, 5, 1, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0007","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b2449d84b03238a0236e0512dc43f3929bd380bb42babe7b1e92cf5fd09f1b7","family_id":"qf-app21-orbit-fixed-audit","seed":202610213007,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":-1,\"b\":3,\"modulus\":5,\"x\":1}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=-1, b=3. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(5, -1, 3, 1, 2, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, -1, 3, 1, 2, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0008","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"84af7b589469cec04b774220d074940955dec5b9a73b4fc94f6a7442375bfe90","family_id":"qf-app21-orbit-fixed-audit","seed":202610213008,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":0,\"b\":5,\"modulus\":6,\"x\":1}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=0, b=5. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(6, 0, 5, 1, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, 0, 5, 1, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0009","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96ec84097dc9f34823156b1cd5cb4b59fd050c09219e0b5046e4ffcf9da9b818","family_id":"qf-app21-orbit-fixed-audit","seed":202610213009,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":1,\"b\":7,\"modulus\":7,\"x\":1}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=1, b=7. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(7, 1, 7, 1, 1, Yes)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, 1, 7, 1, 1, Yes)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0010","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9bbaeb9bc99944e0de4e60cb038031197f4327ef3c30b19f205906a5221c3671","family_id":"qf-app21-orbit-fixed-audit","seed":202610213010,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":2,\"b\":9,\"modulus\":8,\"x\":1}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=2, b=9. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(8, 2, 9, 1, 3, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, 2, 9, 1, 3, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0011","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"793462e4fe1bbfbbdbdbf05ec3a568527da3800d1685697d2d38f490bd4ca03c","family_id":"qf-app21-orbit-fixed-audit","seed":202610213011,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":3,\"b\":11,\"modulus\":9,\"x\":1}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=3, b=11. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(9, 3, 11, 1, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, 3, 11, 1, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0012","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"47c5ee1317ec690301cef34990afcc49355f04b842696c671dd39f47c9d8d7ae","family_id":"qf-app21-orbit-fixed-audit","seed":202610213012,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":4,\"b\":13,\"modulus\":10,\"x\":1}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=4, b=13. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(10, 4, 13, 1, 7, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 4, 13, 1, 7, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0013","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c4d3a32b461df6f61d5a39a85870e6fad48a30d30556d36f9a63cc22a579f7f","family_id":"qf-app21-orbit-fixed-audit","seed":202610213013,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":5,\"b\":15,\"modulus\":5,\"x\":2}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=5, b=15. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=2. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(5, 5, 15, 2, 0, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 5, 15, 2, 0, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0014","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13b3abc258d43392b4abbd88be292c7c24a3cfebd32bc702ccad3c5dbe07fc57","family_id":"qf-app21-orbit-fixed-audit","seed":202610213014,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":6,\"b\":17,\"modulus\":6,\"x\":1}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=6, b=17. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(6, 6, 17, 1, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, 6, 17, 1, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0015","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e43ebe63f11bb386276e50c8b1be3347f55d9e9c54ca15fe47a31a81b42219c","family_id":"qf-app21-orbit-fixed-audit","seed":202610213015,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":7,\"b\":19,\"modulus\":7,\"x\":0}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=7, b=19. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=0. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(7, 7, 19, 0, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, 7, 19, 0, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0016","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"758eb69735e1911b33e9ef5db373636646889672af7ac314d3dda568def90402","family_id":"qf-app21-orbit-fixed-audit","seed":202610213016,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":8,\"b\":21,\"modulus\":8,\"x\":7}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=8, b=21. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=7. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(8, 8, 21, 7, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, 8, 21, 7, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0017","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f9eab884076bf2c6b0ddbd7c46027ddaaca047af4562e9a774cddd6d52b386b","family_id":"qf-app21-orbit-fixed-audit","seed":202610213017,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":9,\"b\":23,\"modulus\":9,\"x\":7}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=9, b=23. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=7. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(9, 9, 23, 7, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, 9, 23, 7, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0018","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a568483636b882295a2763d821ff28f2972c99e05eb58bc0aad69cd8bee37c24","family_id":"qf-app21-orbit-fixed-audit","seed":202610213018,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":10,\"b\":25,\"modulus\":10,\"x\":7}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=10, b=25. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=7. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(10, 10, 25, 7, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 10, 25, 7, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0019","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3adcd7fd2ee44e992d7e0274677e1a6acec4e4f29662a4d36099f3369edb933","family_id":"qf-app21-orbit-fixed-audit","seed":202610213019,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":11,\"b\":27,\"modulus\":5,\"x\":3}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=11, b=27. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(5, 11, 27, 3, 0, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 11, 27, 3, 0, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0020","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0471f3359971d9dd96126dcd7192f6f897b03720225a278a7183b6243ba6dea5","family_id":"qf-app21-orbit-fixed-audit","seed":202610213020,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":12,\"b\":29,\"modulus\":6,\"x\":1}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=12, b=29. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(6, 12, 29, 1, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, 12, 29, 1, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0021","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e00bdd952d6f4dd56905ba3ee2207d0715a2e871beedf038b25c6db482c6100b","family_id":"qf-app21-orbit-fixed-audit","seed":202610213021,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":13,\"b\":31,\"modulus\":7,\"x\":6}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=13, b=31. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=6. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(7, 13, 31, 6, 4, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, 13, 31, 6, 4, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0022","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd9df6ed0581a33c166ecabaf6e00909c90ae9d2fe10297095d84301c2554be9","family_id":"qf-app21-orbit-fixed-audit","seed":202610213022,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":14,\"b\":33,\"modulus\":8,\"x\":5}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=14, b=33. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=5. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(8, 14, 33, 5, 7, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, 14, 33, 5, 7, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0023","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a38eca8e291da0e14ba7e8b14946a28b0db9567a84492c2c02faaaf3d11461ad","family_id":"qf-app21-orbit-fixed-audit","seed":202610213023,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":15,\"b\":35,\"modulus\":9,\"x\":4}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=15, b=35. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=4. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(9, 15, 35, 4, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, 15, 35, 4, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0024","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"243307f5aafd8b9be996a2c400d6f385c22e0f1c24f96fdf47cd0cd2cfb1786a","family_id":"qf-app21-orbit-fixed-audit","seed":202610213024,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":16,\"b\":37,\"modulus\":10,\"x\":3}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=16, b=37. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(10, 16, 37, 3, 5, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 16, 37, 3, 5, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0025","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff29e8c9f459b5c174c38143e001b922fc36295d1c872b158ff66b1b07145a4f","family_id":"qf-app21-orbit-fixed-audit","seed":202610213025,"semantic_template":"qf-app21-orbit-fixed-audit/{\"a\":17,\"b\":39,\"modulus\":5,\"x\":4}"},"fields":{"title":"One State’s Fixed-Point Condition","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=17, b=39. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=4. Give (modulus,a,b,x,T(x),T(x)=x?).","answer":{"kind":"expression","value":"(5, 17, 39, 4, 2, No)","explanation":"Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 17, 39, 4, 2, No)."},"skills":["One State’s Fixed-Point Condition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0026","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1df6f5da79db6f4a1787a7a45ea8388f555519f600cedcaf5a22197776a743ed","family_id":"qf-app21-orbit-return-time-audit","seed":202610213026,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-3,\"b\":-8,\"modulus\":5,\"p\":1,\"x\":0}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=-3, b=-8. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=0, proposed positive return time p=1. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(5, -3, -8, 0, 1, 2, No, 4, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, -3, -8, 0, 1, 2, No, 4, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0027","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69b6d4d4e76e8aa67382508058995c5553c0fc1ca071f4e6b31a6686dc4c9541","family_id":"qf-app21-orbit-return-time-audit","seed":202610213027,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-2,\"b\":-7,\"modulus\":6,\"p\":2,\"x\":1}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=-2, b=-7. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1, proposed positive return time p=2. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(6, -2, -7, 1, 2, 5, No, 3, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, -2, -7, 1, 2, 5, No, 3, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0028","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e18906bbe0cfc6d146960b7882137ce4f97890b5e48efabb62625de83a2b08e9","family_id":"qf-app21-orbit-return-time-audit","seed":202610213028,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-1,\"b\":-6,\"modulus\":7,\"p\":3,\"x\":2}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-1, b=-6. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=2, proposed positive return time p=3. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(7, -1, -6, 2, 3, 6, No, 2, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -1, -6, 2, 3, 6, No, 2, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0029","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3162bfc14c45fc6be8b9b7ca4bc4d9a29ec03ef139e03544a786f50b9ca677a","family_id":"qf-app21-orbit-return-time-audit","seed":202610213029,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":0,\"b\":-5,\"modulus\":8,\"p\":4,\"x\":3}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=0, b=-5. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3, proposed positive return time p=4. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(8, 0, -5, 3, 4, 3, Yes, 1, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, 0, -5, 3, 4, 3, Yes, 1, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0030","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ce1e369ab46b20c2d28f91ac90463e759a0b9b4a430d377f69c04fda03ae634","family_id":"qf-app21-orbit-return-time-audit","seed":202610213030,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":1,\"b\":-4,\"modulus\":9,\"p\":5,\"x\":4}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=1, b=-4. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=4, proposed positive return time p=5. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(9, 1, -4, 4, 5, 2, No, 9, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, 1, -4, 4, 5, 2, No, 9, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0031","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd50eb6a29a129b129b9a17304a39b683304468bab5246f6d256d18cf35366e2","family_id":"qf-app21-orbit-return-time-audit","seed":202610213031,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":2,\"b\":-3,\"modulus\":10,\"p\":6,\"x\":5}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=2, b=-3. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=5, proposed positive return time p=6. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(10, 2, -3, 5, 6, 1, No, 4, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 2, -3, 5, 6, 1, No, 4, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0032","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dacb154fe5d12838e3910d80fefb6a50d88cbc20f05b79bc3e7620b0511f956f","family_id":"qf-app21-orbit-return-time-audit","seed":202610213032,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":3,\"b\":-2,\"modulus\":5,\"p\":7,\"x\":1}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=3, b=-2. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1, proposed positive return time p=7. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(5, 3, -2, 1, 7, 1, Yes, 1, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 3, -2, 1, 7, 1, Yes, 1, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0033","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f89f25d70cf4bd1e8cba52c24a157b159e58af478eeabfe8ecc7098ef7d868d","family_id":"qf-app21-orbit-return-time-audit","seed":202610213033,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-3,\"b\":-1,\"modulus\":6,\"p\":8,\"x\":1}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=-3, b=-1. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1, proposed positive return time p=8. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(6, -3, -1, 1, 8, 5, No, null, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, -3, -1, 1, 8, 5, No, null, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0034","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c0f518fbaa60826dd5a518c808e815be3c5f1b977cf2ba2d020770bb0cbad483","family_id":"qf-app21-orbit-return-time-audit","seed":202610213034,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-2,\"b\":0,\"modulus\":7,\"p\":1,\"x\":1}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-2, b=0. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1, proposed positive return time p=1. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(7, -2, 0, 1, 1, 5, No, 6, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -2, 0, 1, 1, 5, No, 6, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0035","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e78234726b97d68f926131be7aac88f5be6ab5f03ad08d3b2fc1844b24ed938d","family_id":"qf-app21-orbit-return-time-audit","seed":202610213035,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-1,\"b\":1,\"modulus\":8,\"p\":2,\"x\":1}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-1, b=1. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1, proposed positive return time p=2. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(8, -1, 1, 1, 2, 1, Yes, 2, Yes)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -1, 1, 1, 2, 1, Yes, 2, Yes)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0036","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd7089f687c68ddbc50f223cac17335bdd4d2be480043af814a27b9d9f754f2f","family_id":"qf-app21-orbit-return-time-audit","seed":202610213036,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":0,\"b\":2,\"modulus\":9,\"p\":3,\"x\":1}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=0, b=2. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1, proposed positive return time p=3. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(9, 0, 2, 1, 3, 2, No, null, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, 0, 2, 1, 3, 2, No, null, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0037","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ecf4821a530f23528782257c0d3cf8e801176232c83c34de5f6a2cd58079222","family_id":"qf-app21-orbit-return-time-audit","seed":202610213037,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":1,\"b\":3,\"modulus\":10,\"p\":4,\"x\":1}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=1, b=3. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1, proposed positive return time p=4. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(10, 1, 3, 1, 4, 3, No, 10, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 1, 3, 1, 4, 3, No, 10, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0038","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6bf74a5947a9096b990a46804be95c4749194ee5933eea2a4563fed423957b04","family_id":"qf-app21-orbit-return-time-audit","seed":202610213038,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":2,\"b\":4,\"modulus\":5,\"p\":5,\"x\":2}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=2, b=4. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=2, proposed positive return time p=5. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(5, 2, 4, 2, 5, 3, No, 4, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 2, 4, 2, 5, 3, No, 4, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0039","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3296b026ba62151e57153ff3009ba0ba517463751e5052672764536c1de65188","family_id":"qf-app21-orbit-return-time-audit","seed":202610213039,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":3,\"b\":5,\"modulus\":6,\"p\":6,\"x\":1}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=3, b=5. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1, proposed positive return time p=6. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(6, 3, 5, 1, 6, 5, No, null, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, 3, 5, 1, 6, 5, No, null, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0040","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c0108342d4d3d96c99ad0c15cb4a051c5dffc21e3cb21d18a883271140d34e65","family_id":"qf-app21-orbit-return-time-audit","seed":202610213040,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-3,\"b\":6,\"modulus\":7,\"p\":7,\"x\":0}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-3, b=6. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=0, proposed positive return time p=7. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(7, -3, 6, 0, 7, 6, No, 3, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -3, 6, 0, 7, 6, No, 3, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0041","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b27e7be78320dafb1978c507ef1b8397781d1bead125722062cc48c615f51bc","family_id":"qf-app21-orbit-return-time-audit","seed":202610213041,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-2,\"b\":7,\"modulus\":8,\"p\":8,\"x\":7}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-2, b=7. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=7, proposed positive return time p=8. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(8, -2, 7, 7, 8, 5, No, null, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -2, 7, 7, 8, 5, No, null, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0042","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"758bb68e54bcb24501353915c22764c01878a2f322dfdd47cac4c37282d3503f","family_id":"qf-app21-orbit-return-time-audit","seed":202610213042,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-1,\"b\":8,\"modulus\":9,\"p\":1,\"x\":7}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=-1, b=8. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=7, proposed positive return time p=1. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(9, -1, 8, 7, 1, 1, No, 2, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, -1, 8, 7, 1, 1, No, 2, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0043","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9161b072817ad0ab08c31131eb7fe663ae67a3640f2b1c055756fe57f9eb8ff7","family_id":"qf-app21-orbit-return-time-audit","seed":202610213043,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":0,\"b\":9,\"modulus\":10,\"p\":2,\"x\":7}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=0, b=9. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=7, proposed positive return time p=2. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(10, 0, 9, 7, 2, 9, No, null, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 0, 9, 7, 2, 9, No, null, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0044","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37ff3c1c9d5cf83246627241dda2677736e7e15bd8d8301475277b3aa84aedec","family_id":"qf-app21-orbit-return-time-audit","seed":202610213044,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":1,\"b\":10,\"modulus\":5,\"p\":3,\"x\":3}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=1, b=10. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3, proposed positive return time p=3. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(5, 1, 10, 3, 3, 3, Yes, 1, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 1, 10, 3, 3, 3, Yes, 1, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0045","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b389c21d7946d736e4ae5c9084064b8a2b71864691a376e991cfa7b83eb6335","family_id":"qf-app21-orbit-return-time-audit","seed":202610213045,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":2,\"b\":11,\"modulus\":6,\"p\":4,\"x\":1}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=2, b=11. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1, proposed positive return time p=4. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(6, 2, 11, 1, 4, 1, Yes, 1, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, 2, 11, 1, 4, 1, Yes, 1, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0046","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d4099c6c1f4f14f7ccf7a88044f8e8867a84445b9c0674987e411c51fdc71f48","family_id":"qf-app21-orbit-return-time-audit","seed":202610213046,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":3,\"b\":12,\"modulus\":7,\"p\":5,\"x\":6}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=3, b=12. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=6, proposed positive return time p=5. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(7, 3, 12, 6, 5, 5, No, 6, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, 3, 12, 6, 5, 5, No, 6, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0047","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01a1dad18a44336e95819a7c1a345b3bd19ac27b8ca6aa0330402d02d14bffde","family_id":"qf-app21-orbit-return-time-audit","seed":202610213047,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-3,\"b\":13,\"modulus\":8,\"p\":6,\"x\":5}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-3, b=13. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=5, proposed positive return time p=6. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(8, -3, 13, 5, 6, 7, No, 8, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -3, 13, 5, 6, 7, No, 8, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0048","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e9c2e184af94c5f7da0f09b1edbf239e05a050a652a732dce7726f308997fdd","family_id":"qf-app21-orbit-return-time-audit","seed":202610213048,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-2,\"b\":14,\"modulus\":9,\"p\":7,\"x\":4}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=-2, b=14. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=4, proposed positive return time p=7. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(9, -2, 14, 4, 7, 0, No, 9, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, -2, 14, 4, 7, 0, No, 9, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0049","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5b1a1be4419dbeea22f5c9d1272da75eb09e036ca6a20483d7f762f4c2a8c88","family_id":"qf-app21-orbit-return-time-audit","seed":202610213049,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":-1,\"b\":15,\"modulus\":10,\"p\":8,\"x\":3}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=-1, b=15. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3, proposed positive return time p=8. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(10, -1, 15, 3, 8, 3, Yes, 2, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, -1, 15, 3, 8, 3, Yes, 2, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0050","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"117860c53c6cb4d671644da9491327c94c07d83f458a1a2d4b99529a723a1191","family_id":"qf-app21-orbit-return-time-audit","seed":202610213050,"semantic_template":"qf-app21-orbit-return-time-audit/{\"a\":0,\"b\":16,\"modulus\":5,\"p\":1,\"x\":4}"},"fields":{"title":"A Return Time Need Not Be the Least Positive Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=0, b=16. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=4, proposed positive return time p=1. Give (modulus,a,b,x,p,T^p(x),returns to x?,least positive period or null if x is not periodic,is the least period exactly p?).","answer":{"kind":"expression","value":"(5, 0, 16, 4, 1, 1, No, null, No)","explanation":"Iterate p times, then separately enumerate to the first repeat to determine the least period. Entering a later cycle does not make the starting point periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 0, 16, 4, 1, 1, No, null, No)."},"skills":["A Return Time Need Not Be the Least Positive Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0051","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca83b5ebd6af4b4366b6804964a8c7f6a1ca2b1be82733880d77962451f4a020","family_id":"qf-app21-orbit-least-period","seed":202610213051,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-3,\"b\":-8,\"modulus\":5,\"x\":1}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=-3, b=-8. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(5, -3, -8, 1, [1, 4, 0, 2], 1, 0, 4)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, -3, -8, 1, [1, 4, 0, 2], 1, 0, 4)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0052","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a8e456d26047d81dd1780fc0eea8b3339b4dbfe18d0479f4840ac65b4add346","family_id":"qf-app21-orbit-least-period","seed":202610213052,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-2,\"b\":-7,\"modulus\":6,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=-2, b=-7. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(6, -2, -7, 3, [3, 5, 1], 3, 0, 3)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, -2, -7, 3, [3, 5, 1], 3, 0, 3)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0053","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f915e6db702a8e04259af656bf13e2be823e1344b88bd88ed3bde44f78c39ef","family_id":"qf-app21-orbit-least-period","seed":202610213053,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-1,\"b\":-6,\"modulus\":7,\"x\":5}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-1, b=-6. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=5. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(7, -1, -6, 5, [5, 3], 5, 0, 2)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -1, -6, 5, [5, 3], 5, 0, 2)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0054","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8689bb0c4ddf8d46fe6a68c7d163a4f2af9883860197b94e2d433d8f6c986938","family_id":"qf-app21-orbit-least-period","seed":202610213054,"semantic_template":"qf-app21-orbit-least-period/{\"a\":0,\"b\":-5,\"modulus\":8,\"x\":7}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=0, b=-5. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=7. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(8, 0, -5, 7, [7, 3], 3, 1, nonperiodic-point)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, 0, -5, 7, [7, 3], 3, 1, nonperiodic-point)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0055","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cad87d2d65017ee1380faed41d8086bce21a5412b77fd39ef833e7327232064f","family_id":"qf-app21-orbit-least-period","seed":202610213055,"semantic_template":"qf-app21-orbit-least-period/{\"a\":1,\"b\":-4,\"modulus\":9,\"x\":0}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=1, b=-4. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=0. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(9, 1, -4, 0, [0, 5, 1, 6, 2, 7, 3, 8, 4], 0, 0, 9)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, 1, -4, 0, [0, 5, 1, 6, 2, 7, 3, 8, 4], 0, 0, 9)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0056","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7e17a306de59b2f8860376c242cfbb712453ead36256059aaf0d81b87db4712f","family_id":"qf-app21-orbit-least-period","seed":202610213056,"semantic_template":"qf-app21-orbit-least-period/{\"a\":2,\"b\":-3,\"modulus\":10,\"x\":1}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=2, b=-3. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(10, 2, -3, 1, [1, 9, 5, 7], 1, 0, 4)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 2, -3, 1, [1, 9, 5, 7], 1, 0, 4)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0057","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db24152ad0a4400b8fb37c8106960791b9edb38bd6da67ec2e1c84b909e01ffa","family_id":"qf-app21-orbit-least-period","seed":202610213057,"semantic_template":"qf-app21-orbit-least-period/{\"a\":3,\"b\":-2,\"modulus\":5,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=3, b=-2. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(5, 3, -2, 3, [3, 2, 4, 0], 3, 0, 4)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 3, -2, 3, [3, 2, 4, 0], 3, 0, 4)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0058","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"167585c1f65037ec6dc4b8a89533aedb249efa77072bfd993273bac275b12c17","family_id":"qf-app21-orbit-least-period","seed":202610213058,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-3,\"b\":-1,\"modulus\":6,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=-3, b=-1. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(6, -3, -1, 3, [3, 2, 5], 2, 1, nonperiodic-point)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, -3, -1, 3, [3, 2, 5], 2, 1, nonperiodic-point)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0059","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dac5bd747cda8d279f1c1a2bf7115da6893d4a8a985a2e4317a69d78e039d0e1","family_id":"qf-app21-orbit-least-period","seed":202610213059,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-2,\"b\":0,\"modulus\":7,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-2, b=0. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(7, -2, 0, 3, [3, 1, 5, 4, 6, 2], 3, 0, 6)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -2, 0, 3, [3, 1, 5, 4, 6, 2], 3, 0, 6)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0060","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22614d9c42afd59f8628256135e8794a9eeb06f9ec5c9a041811848f54ce5be2","family_id":"qf-app21-orbit-least-period","seed":202610213060,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-1,\"b\":1,\"modulus\":8,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-1, b=1. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(8, -1, 1, 3, [3, 6], 3, 0, 2)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -1, 1, 3, [3, 6], 3, 0, 2)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0061","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ff226dd34a3e69bc0d0475e05dac206453f08d0d2bb6692c805df5b9ed6c519","family_id":"qf-app21-orbit-least-period","seed":202610213061,"semantic_template":"qf-app21-orbit-least-period/{\"a\":0,\"b\":2,\"modulus\":9,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=0, b=2. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(9, 0, 2, 3, [3, 2], 2, 1, nonperiodic-point)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, 0, 2, 3, [3, 2], 2, 1, nonperiodic-point)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0062","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ef4f7ad46b99de89e12b8dbfee194e404a2b4016e09b1ad3418dd0795c1a1b1","family_id":"qf-app21-orbit-least-period","seed":202610213062,"semantic_template":"qf-app21-orbit-least-period/{\"a\":1,\"b\":3,\"modulus\":10,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=1, b=3. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(10, 1, 3, 3, [3, 6, 9, 2, 5, 8, 1, 4, 7, 0], 3, 0, 10)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 1, 3, 3, [3, 6, 9, 2, 5, 8, 1, 4, 7, 0], 3, 0, 10)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0063","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec2efca324d97ce1c1fada6ff1c7c17ba5ac17ec0ad07b4fd3b1bea6c1119664","family_id":"qf-app21-orbit-least-period","seed":202610213063,"semantic_template":"qf-app21-orbit-least-period/{\"a\":2,\"b\":4,\"modulus\":5,\"x\":0}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=2, b=4. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=0. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(5, 2, 4, 0, [0, 4, 2, 3], 0, 0, 4)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 2, 4, 0, [0, 4, 2, 3], 0, 0, 4)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0064","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46629f3a005422a0b3ce10187bb09b831b07cca0c728ae6f8e17103f150d65d3","family_id":"qf-app21-orbit-least-period","seed":202610213064,"semantic_template":"qf-app21-orbit-least-period/{\"a\":3,\"b\":5,\"modulus\":6,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=3, b=5. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(6, 3, 5, 3, [3, 2, 5], 2, 1, nonperiodic-point)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, 3, 5, 3, [3, 2, 5], 2, 1, nonperiodic-point)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0065","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e4150086bdf4f5e3e5550feb566429237c7d1ef9728cdee886770a9536b11aa","family_id":"qf-app21-orbit-least-period","seed":202610213065,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-3,\"b\":6,\"modulus\":7,\"x\":1}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-3, b=6. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(7, -3, 6, 1, [1, 3, 4], 1, 0, 3)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -3, 6, 1, [1, 3, 4], 1, 0, 3)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0066","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5bd849902c0e8c938fe3a1e844258ae68ed5b3e0f0f160a8a35ccbd94272989","family_id":"qf-app21-orbit-least-period","seed":202610213066,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-2,\"b\":7,\"modulus\":8,\"x\":7}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-2, b=7. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=7. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(8, -2, 7, 7, [7, 1, 5], 5, 2, nonperiodic-point)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -2, 7, 7, [7, 1, 5], 5, 2, nonperiodic-point)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0067","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1fd9eac82d574536354a40cd8be9c133969e5fdf0383fd6dd5c004d69edf7a90","family_id":"qf-app21-orbit-least-period","seed":202610213067,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-1,\"b\":8,\"modulus\":9,\"x\":6}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=-1, b=8. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=6. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(9, -1, 8, 6, [6, 2], 6, 0, 2)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, -1, 8, 6, [6, 2], 6, 0, 2)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0068","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"776cd2cc696c4c2c4765a3ea28f86e8837c112b75e4bbb52a54fc54d42814da4","family_id":"qf-app21-orbit-least-period","seed":202610213068,"semantic_template":"qf-app21-orbit-least-period/{\"a\":0,\"b\":9,\"modulus\":10,\"x\":5}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=0, b=9. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=5. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(10, 0, 9, 5, [5, 9], 9, 1, nonperiodic-point)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 0, 9, 5, [5, 9], 9, 1, nonperiodic-point)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0069","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d2ec7303f6625574e7c494254017ef827bd683530b62058eb85f9107653c76b7","family_id":"qf-app21-orbit-least-period","seed":202610213069,"semantic_template":"qf-app21-orbit-least-period/{\"a\":1,\"b\":10,\"modulus\":5,\"x\":2}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=1, b=10. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=2. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(5, 1, 10, 2, [2], 2, 0, 1)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 1, 10, 2, [2], 2, 0, 1)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0070","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d130677bbd5b80ce4dcb83fe4e837d67667188a7c8fa09cdf33bef5755e2bb9c","family_id":"qf-app21-orbit-least-period","seed":202610213070,"semantic_template":"qf-app21-orbit-least-period/{\"a\":2,\"b\":11,\"modulus\":6,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=2, b=11. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(6, 2, 11, 3, [3, 5], 3, 0, 2)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, 2, 11, 3, [3, 5], 3, 0, 2)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0071","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d6f672bec48aeb9ec6150de02fe0803b9bc724b5e0f81395259aa746068e8b9","family_id":"qf-app21-orbit-least-period","seed":202610213071,"semantic_template":"qf-app21-orbit-least-period/{\"a\":3,\"b\":12,\"modulus\":7,\"x\":6}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=3, b=12. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=6. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(7, 3, 12, 6, [6, 2, 4, 3, 0, 5], 6, 0, 6)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, 3, 12, 6, [6, 2, 4, 3, 0, 5], 6, 0, 6)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0072","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d3415a0de988e5469015ca608e953c3fc6b32a49c4cf762136a8534f39e8ed0","family_id":"qf-app21-orbit-least-period","seed":202610213072,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-3,\"b\":13,\"modulus\":8,\"x\":3}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-3, b=13. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=3. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(8, -3, 13, 3, [3, 4, 1, 2, 7, 0, 5, 6], 3, 0, 8)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -3, 13, 3, [3, 4, 1, 2, 7, 0, 5, 6], 3, 0, 8)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0073","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17493151dca35e5cc3456e2223045d500bed3ebce2de949b15deb76046aa6168","family_id":"qf-app21-orbit-least-period","seed":202610213073,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-2,\"b\":14,\"modulus\":9,\"x\":0}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=-2, b=14. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=0. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(9, -2, 14, 0, [0, 5, 4, 6, 2, 1, 3, 8, 7], 0, 0, 9)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, -2, 14, 0, [0, 5, 4, 6, 2, 1, 3, 8, 7], 0, 0, 9)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0074","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25fcdf2102487e4a7a853456ee0da5279f1ae7888eff0044e35cc11099e8ebde","family_id":"qf-app21-orbit-least-period","seed":202610213074,"semantic_template":"qf-app21-orbit-least-period/{\"a\":-1,\"b\":15,\"modulus\":10,\"x\":7}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=-1, b=15. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=7. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(10, -1, 15, 7, [7, 8], 7, 0, 2)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, -1, 15, 7, [7, 8], 7, 0, 2)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0075","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef73a1abc5fa2636c6c1dd5e2d27a602b7f689223a3243caa7a6db3974f2e8b0","family_id":"qf-app21-orbit-least-period","seed":202610213075,"semantic_template":"qf-app21-orbit-least-period/{\"a\":0,\"b\":16,\"modulus\":5,\"x\":4}"},"fields":{"title":"First Repeat, Transient Length and the Starting Point’s Period","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=0, b=16. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=4. Give (modulus,a,b,x,trace before first repeat,repeated state,zero-based cycle-start index,least positive period or nonperiodic-point).","answer":{"kind":"expression","value":"(5, 0, 16, 4, [4, 1], 1, 1, nonperiodic-point)","explanation":"Stop immediately before appending the first repeated state. Only cycle-start index 0 makes the original point periodic; otherwise it is eventually periodic but not itself periodic. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 0, 16, 4, [4, 1], 1, 1, nonperiodic-point)."},"skills":["First Repeat, Transient Length and the Starting Point’s Period"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0076","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"67cbb86ed17bb803ba2a39d2fc105d7aedcacffd0848a80afe494993961cb6ae","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213076,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-3,\"b\":-8,\"modulus\":5,\"period_bound\":1}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=-3, b=-8. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=1. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(5, -3, -8, 1, [[0, 4], [1, 4], [2, 4], [3, 1], [4, 4]], [3])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, -3, -8, 1, [[0, 4], [1, 4], [2, 4], [3, 1], [4, 4]], [3])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0077","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37f6c809c6737b24c878391d702472de767ff9c8ab704c60d3743833ee8217ac","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213077,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-2,\"b\":-7,\"modulus\":6,\"period_bound\":2}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=-2, b=-7. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=2. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(6, -2, -7, 2, [[0, nonperiodic-point], [1, 3], [2, nonperiodic-point], [3, 3], [4, nonperiodic-point], [5, 3]], [])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, -2, -7, 2, [[0, nonperiodic-point], [1, 3], [2, nonperiodic-point], [3, 3], [4, nonperiodic-point], [5, 3]], [])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0078","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2dd1f3532faf84930044f0bd5ef9bacf0d9db09a8fc7b34c5791f5641767225c","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213078,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-1,\"b\":-6,\"modulus\":7,\"period_bound\":3}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-1, b=-6. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=3. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(7, -1, -6, 3, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 1], [5, 2], [6, 2]], [0, 1, 2, 3, 4, 5, 6])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -1, -6, 3, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 1], [5, 2], [6, 2]], [0, 1, 2, 3, 4, 5, 6])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0079","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9133f1cc43b8734bc341aa03aae55b28043ed237bcb4970d56ddd07306603718","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213079,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":0,\"b\":-5,\"modulus\":8,\"period_bound\":4}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=0, b=-5. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=4. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(8, 0, -5, 4, [[0, nonperiodic-point], [1, nonperiodic-point], [2, nonperiodic-point], [3, 1], [4, nonperiodic-point], [5, nonperiodic-point], [6, nonperiodic-point], [7, nonperiodic-point]], [3])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, 0, -5, 4, [[0, nonperiodic-point], [1, nonperiodic-point], [2, nonperiodic-point], [3, 1], [4, nonperiodic-point], [5, nonperiodic-point], [6, nonperiodic-point], [7, nonperiodic-point]], [3])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0080","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66a82dbcee34b93742ef8236494393772782c1874b9d91661ef5da6a3701f944","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213080,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":1,\"b\":-4,\"modulus\":9,\"period_bound\":5}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=1, b=-4. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=5. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(9, 1, -4, 5, [[0, 9], [1, 9], [2, 9], [3, 9], [4, 9], [5, 9], [6, 9], [7, 9], [8, 9]], [])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, 1, -4, 5, [[0, 9], [1, 9], [2, 9], [3, 9], [4, 9], [5, 9], [6, 9], [7, 9], [8, 9]], [])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0081","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9ee1ff35e53b7bc88c529786c3e1ab6299dc5462903c5e4b9210ca51c8372cb","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213081,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":2,\"b\":-3,\"modulus\":10,\"period_bound\":6}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=2, b=-3. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=6. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(10, 2, -3, 6, [[0, nonperiodic-point], [1, 4], [2, nonperiodic-point], [3, 1], [4, nonperiodic-point], [5, 4], [6, nonperiodic-point], [7, 4], [8, nonperiodic-point], [9, 4]], [1, 3, 5, 7, 9])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 2, -3, 6, [[0, nonperiodic-point], [1, 4], [2, nonperiodic-point], [3, 1], [4, nonperiodic-point], [5, 4], [6, nonperiodic-point], [7, 4], [8, nonperiodic-point], [9, 4]], [1, 3, 5, 7, 9])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0082","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2187163fec0d6c806e18c379d42db0a6d88f24af8c4889fa0a2254fe4d76e9a5","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213082,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":3,\"b\":-2,\"modulus\":5,\"period_bound\":1}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=3, b=-2. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=1. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(5, 3, -2, 1, [[0, 4], [1, 1], [2, 4], [3, 4], [4, 4]], [1])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 3, -2, 1, [[0, 4], [1, 1], [2, 4], [3, 4], [4, 4]], [1])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0083","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7288104c515a74ee0355484fcf62e4125820f484663f5bdbbd48ecdf1ab7a54e","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213083,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-3,\"b\":-1,\"modulus\":6,\"period_bound\":2}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=-3, b=-1. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=2. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(6, -3, -1, 2, [[0, nonperiodic-point], [1, nonperiodic-point], [2, 2], [3, nonperiodic-point], [4, nonperiodic-point], [5, 2]], [2, 5])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, -3, -1, 2, [[0, nonperiodic-point], [1, nonperiodic-point], [2, 2], [3, nonperiodic-point], [4, nonperiodic-point], [5, 2]], [2, 5])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0084","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f2d35162e33399d870babde5e6bd5449358cb2c299390b29553070473ef5f8f","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213084,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-2,\"b\":0,\"modulus\":7,\"period_bound\":3}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-2, b=0. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=3. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(7, -2, 0, 3, [[0, 1], [1, 6], [2, 6], [3, 6], [4, 6], [5, 6], [6, 6]], [0])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -2, 0, 3, [[0, 1], [1, 6], [2, 6], [3, 6], [4, 6], [5, 6], [6, 6]], [0])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0085","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a9799b2341005ebb14d03986bb20fb5dd056c22cdc45dec07a72878ba443491","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213085,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-1,\"b\":1,\"modulus\":8,\"period_bound\":4}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-1, b=1. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=4. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(8, -1, 1, 4, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2]], [0, 1, 2, 3, 4, 5, 6, 7])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -1, 1, 4, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2]], [0, 1, 2, 3, 4, 5, 6, 7])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0086","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7c13fb268d7ac5fee567f9c671af989a76d35b803f147f5025a3a318cb4b81a","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213086,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":0,\"b\":2,\"modulus\":9,\"period_bound\":5}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=0, b=2. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=5. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(9, 0, 2, 5, [[0, nonperiodic-point], [1, nonperiodic-point], [2, 1], [3, nonperiodic-point], [4, nonperiodic-point], [5, nonperiodic-point], [6, nonperiodic-point], [7, nonperiodic-point], [8, nonperiodic-point]], [2])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, 0, 2, 5, [[0, nonperiodic-point], [1, nonperiodic-point], [2, 1], [3, nonperiodic-point], [4, nonperiodic-point], [5, nonperiodic-point], [6, nonperiodic-point], [7, nonperiodic-point], [8, nonperiodic-point]], [2])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0087","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"407fd52f297928c03590b57c01edb9d5e2cae2a7639a36a108934f7afe6f2c4a","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213087,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":1,\"b\":3,\"modulus\":10,\"period_bound\":6}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=1, b=3. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=6. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(10, 1, 3, 6, [[0, 10], [1, 10], [2, 10], [3, 10], [4, 10], [5, 10], [6, 10], [7, 10], [8, 10], [9, 10]], [])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 1, 3, 6, [[0, 10], [1, 10], [2, 10], [3, 10], [4, 10], [5, 10], [6, 10], [7, 10], [8, 10], [9, 10]], [])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0088","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5236dd304b417a1c8a46a70b74c7f5b9ea394e878796cac5b81a70e3f64abf7","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213088,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":2,\"b\":4,\"modulus\":5,\"period_bound\":1}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=2, b=4. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=1. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(5, 2, 4, 1, [[0, 4], [1, 1], [2, 4], [3, 4], [4, 4]], [1])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 2, 4, 1, [[0, 4], [1, 1], [2, 4], [3, 4], [4, 4]], [1])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0089","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"476fb8df20760e6dc11469c4fe877226111835a1d0f4fc986ce56085694a5840","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213089,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":3,\"b\":5,\"modulus\":6,\"period_bound\":2}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=3, b=5. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=2. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(6, 3, 5, 2, [[0, nonperiodic-point], [1, nonperiodic-point], [2, 2], [3, nonperiodic-point], [4, nonperiodic-point], [5, 2]], [2, 5])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, 3, 5, 2, [[0, nonperiodic-point], [1, nonperiodic-point], [2, 2], [3, nonperiodic-point], [4, nonperiodic-point], [5, 2]], [2, 5])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0090","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe51b68791276032f1fc59a75375d9db2b6bc92eb40b6d337d949c93654cc73e","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213090,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-3,\"b\":6,\"modulus\":7,\"period_bound\":3}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=-3, b=6. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=3. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(7, -3, 6, 3, [[0, 3], [1, 3], [2, 3], [3, 3], [4, 3], [5, 1], [6, 3]], [0, 1, 2, 3, 4, 5, 6])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, -3, 6, 3, [[0, 3], [1, 3], [2, 3], [3, 3], [4, 3], [5, 1], [6, 3]], [0, 1, 2, 3, 4, 5, 6])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0091","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44fd76902e5c09d6d8f7d5b6b509e5ab94934f51e995764a6de9c9cf691fc24d","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213091,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-2,\"b\":7,\"modulus\":8,\"period_bound\":4}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-2, b=7. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=4. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(8, -2, 7, 4, [[0, nonperiodic-point], [1, nonperiodic-point], [2, nonperiodic-point], [3, nonperiodic-point], [4, nonperiodic-point], [5, 1], [6, nonperiodic-point], [7, nonperiodic-point]], [5])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -2, 7, 4, [[0, nonperiodic-point], [1, nonperiodic-point], [2, nonperiodic-point], [3, nonperiodic-point], [4, nonperiodic-point], [5, 1], [6, nonperiodic-point], [7, nonperiodic-point]], [5])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0092","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5072a16342875375bede6550ad31d3de3a9959724407e2abf82c83a8100e1705","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213092,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-1,\"b\":8,\"modulus\":9,\"period_bound\":5}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=-1, b=8. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=5. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(9, -1, 8, 5, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 1], [5, 2], [6, 2], [7, 2], [8, 2]], [0, 1, 2, 3, 4, 5, 6, 7, 8])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, -1, 8, 5, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 1], [5, 2], [6, 2], [7, 2], [8, 2]], [0, 1, 2, 3, 4, 5, 6, 7, 8])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0093","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99b76a5a0e8a7c47d06740422dd8fd8b3bb718fa9d9ade3acecabad9a26b325d","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213093,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":0,\"b\":9,\"modulus\":10,\"period_bound\":6}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=0, b=9. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=6. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(10, 0, 9, 6, [[0, nonperiodic-point], [1, nonperiodic-point], [2, nonperiodic-point], [3, nonperiodic-point], [4, nonperiodic-point], [5, nonperiodic-point], [6, nonperiodic-point], [7, nonperiodic-point], [8, nonperiodic-point], [9, 1]], [9])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 0, 9, 6, [[0, nonperiodic-point], [1, nonperiodic-point], [2, nonperiodic-point], [3, nonperiodic-point], [4, nonperiodic-point], [5, nonperiodic-point], [6, nonperiodic-point], [7, nonperiodic-point], [8, nonperiodic-point], [9, 1]], [9])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0094","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"60d138d85a48383f16e78c860d87b322363239ee66468c431b2a9425a2b3ed5c","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213094,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":1,\"b\":10,\"modulus\":5,\"period_bound\":1}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=1, b=10. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=1. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(5, 1, 10, 1, [[0, 1], [1, 1], [2, 1], [3, 1], [4, 1]], [0, 1, 2, 3, 4])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 1, 10, 1, [[0, 1], [1, 1], [2, 1], [3, 1], [4, 1]], [0, 1, 2, 3, 4])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0095","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b6acc066ee12462fe9f1211f4b57cc4fd657e2f189fd1f30b5398f4ca443c7a","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213095,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":2,\"b\":11,\"modulus\":6,\"period_bound\":2}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(6)={0,…,5}, T(x)=(a*x+b) mod 6, a=2, b=11. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=2. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(6, 2, 11, 2, [[0, nonperiodic-point], [1, 1], [2, nonperiodic-point], [3, 2], [4, nonperiodic-point], [5, 2]], [1, 3, 5])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (6, 2, 11, 2, [[0, nonperiodic-point], [1, 1], [2, nonperiodic-point], [3, 2], [4, nonperiodic-point], [5, 2]], [1, 3, 5])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0096","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b752827620115646f5ba56bd0c2322a3b94a5a459944083465e0038fcf9d90a","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213096,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":3,\"b\":12,\"modulus\":7,\"period_bound\":3}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(7)={0,…,6}, T(x)=(a*x+b) mod 7, a=3, b=12. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=3. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(7, 3, 12, 3, [[0, 6], [1, 1], [2, 6], [3, 6], [4, 6], [5, 6], [6, 6]], [1])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (7, 3, 12, 3, [[0, 6], [1, 1], [2, 6], [3, 6], [4, 6], [5, 6], [6, 6]], [1])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0097","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"695246bfd35a007ec0cf502fa38f3c6253fccdc99eb1c03b12c3a099fc5c4471","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213097,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-3,\"b\":13,\"modulus\":8,\"period_bound\":4}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(8)={0,…,7}, T(x)=(a*x+b) mod 8, a=-3, b=13. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=4. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(8, -3, 13, 4, [[0, 8], [1, 8], [2, 8], [3, 8], [4, 8], [5, 8], [6, 8], [7, 8]], [])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (8, -3, 13, 4, [[0, 8], [1, 8], [2, 8], [3, 8], [4, 8], [5, 8], [6, 8], [7, 8]], [])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0098","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"48d96ad0d62d066aceab727dbad8bd35e1dd24974b8ed62bf3554e40757ab33a","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213098,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-2,\"b\":14,\"modulus\":9,\"period_bound\":5}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(9)={0,…,8}, T(x)=(a*x+b) mod 9, a=-2, b=14. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=5. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(9, -2, 14, 5, [[0, 9], [1, 9], [2, 9], [3, 9], [4, 9], [5, 9], [6, 9], [7, 9], [8, 9]], [])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (9, -2, 14, 5, [[0, 9], [1, 9], [2, 9], [3, 9], [4, 9], [5, 9], [6, 9], [7, 9], [8, 9]], [])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0099","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f16bd1e1649c29c04d4094275ec35ce6345161d1c49c8bed0b6ed5d24e66d3e","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213099,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":-1,\"b\":15,\"modulus\":10,\"period_bound\":6}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=-1, b=15. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=6. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(10, -1, 15, 6, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2]], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, -1, 15, 6, [[0, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [6, 2], [7, 2], [8, 2], [9, 2]], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261021-r03-orbit-fixed-periodic-0100","batch_id":"batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9bc96164be68908d5454330fa9d053cb7c4d92df8dd1f6ae01f133f0e269889","family_id":"qf-app21-orbit-periodic-inventory","seed":202610213100,"semantic_template":"qf-app21-orbit-periodic-inventory/{\"a\":0,\"b\":16,\"modulus\":5,\"period_bound\":1}"},"fields":{"title":"Every Point’s Least Period in the Complete Finite Model","stem":"Use the complete finite self-map on Fin(5)={0,…,4}, T(x)=(a*x+b) mod 5, a=0, b=16. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. Positive period bound=1. Give (modulus,a,b,bound,every [point,least period or nonperiodic-point] row in point order,points whose least positive period is at most the bound).","answer":{"kind":"expression","value":"(5, 0, 16, 1, [[0, nonperiodic-point], [1, 1], [2, nonperiodic-point], [3, nonperiodic-point], [4, nonperiodic-point]], [1])","explanation":"Enumerate every state, distinguish transient starting points from cycle points and filter only defined least periods. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (5, 0, 16, 1, [[0, nonperiodic-point], [1, 1], [2, nonperiodic-point], [3, nonperiodic-point], [4, nonperiodic-point]], [1])."},"skills":["Every Point’s Least Period in the Complete Finite Model"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}