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模n乘法階基礎 · 脈衝 R07
本批已通過 Schema、去重、來源快照與獨立答案重算;仍未標記為人工審定。
本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留;AI 覆核、有限計算與 Lean 形式證明分開標示。
batch-20261015-advance-20261116-r07-multiplicative-order-mod-n-foundations · 2026-10-15 · 100 題
欄序為來源綁定的呈現補充;原始 JSONL 與數學答案不變。
欄序與來源綁定資料
1. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":3,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(3, 2, 2, 1, 2, 是, unit_mod_n_qualified)
Wave6第十層核對=(3, 2, 2, 1, 2, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
2. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":4,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(4, 3, 3, 1, 2, 是, unit_mod_n_qualified)
Wave6第十層核對=(4, 3, 3, 1, 2, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
3. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":5,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(5, 2, 2, 1, 4, 是, unit_mod_n_qualified)
Wave6第十層核對=(5, 2, 2, 1, 4, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
4. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":6,"a":5}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(6, 5, 5, 1, 2, 是, unit_mod_n_qualified)
Wave6第十層核對=(6, 5, 5, 1, 2, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
5. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":7,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(7, 2, 2, 1, 6, 是, unit_mod_n_qualified)
Wave6第十層核對=(7, 2, 2, 1, 6, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
6. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":8,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(8, 3, 3, 1, 4, 是, unit_mod_n_qualified)
Wave6第十層核對=(8, 3, 3, 1, 4, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
7. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":9,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(9, 2, 2, 1, 6, 是, unit_mod_n_qualified)
Wave6第十層核對=(9, 2, 2, 1, 6, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
8. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":10,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(10, 3, 3, 1, 4, 是, unit_mod_n_qualified)
Wave6第十層核對=(10, 3, 3, 1, 4, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
9. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":11,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(11, 2, 2, 1, 10, 是, unit_mod_n_qualified)
Wave6第十層核對=(11, 2, 2, 1, 10, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
10. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":12,"a":5}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(12, 5, 5, 1, 4, 是, unit_mod_n_qualified)
Wave6第十層核對=(12, 5, 5, 1, 4, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
11. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":13,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(13, 2, 2, 1, 12, 是, unit_mod_n_qualified)
Wave6第十層核對=(13, 2, 2, 1, 12, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
12. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":14,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(14, 3, 3, 1, 6, 是, unit_mod_n_qualified)
Wave6第十層核對=(14, 3, 3, 1, 6, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
13. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":15,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(15, 2, 2, 1, 8, 是, unit_mod_n_qualified)
Wave6第十層核對=(15, 2, 2, 1, 8, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
14. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":16,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(16, 3, 3, 1, 8, 是, unit_mod_n_qualified)
Wave6第十層核對=(16, 3, 3, 1, 8, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
15. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":17,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(17, 2, 2, 1, 16, 是, unit_mod_n_qualified)
Wave6第十層核對=(17, 2, 2, 1, 16, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
16. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":18,"a":5}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(18, 5, 5, 1, 6, 是, unit_mod_n_qualified)
Wave6第十層核對=(18, 5, 5, 1, 6, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
17. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":19,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(19, 2, 2, 1, 18, 是, unit_mod_n_qualified)
Wave6第十層核對=(19, 2, 2, 1, 18, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
18. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":20,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(20, 3, 3, 1, 8, 是, unit_mod_n_qualified)
Wave6第十層核對=(20, 3, 3, 1, 8, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
19. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":21,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(21, 2, 2, 1, 12, 是, unit_mod_n_qualified)
Wave6第十層核對=(21, 2, 2, 1, 12, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
20. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":22,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(22, 3, 3, 1, 10, 是, unit_mod_n_qualified)
Wave6第十層核對=(22, 3, 3, 1, 10, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
21. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":23,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(23, 2, 2, 1, 22, 是, unit_mod_n_qualified)
Wave6第十層核對=(23, 2, 2, 1, 22, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
22. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":24,"a":5}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(24, 5, 5, 1, 8, 是, unit_mod_n_qualified)
Wave6第十層核對=(24, 5, 5, 1, 8, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
23. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":25,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(25, 2, 2, 1, 20, 是, unit_mod_n_qualified)
Wave6第十層核對=(25, 2, 2, 1, 20, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
24. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":26,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(26, 3, 3, 1, 12, 是, unit_mod_n_qualified)
Wave6第十層核對=(26, 3, 3, 1, 12, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
25. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification;資料={"n":27,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。
查看答案與解釋
(27, 2, 2, 1, 18, 是, unit_mod_n_qualified)
Wave6第十層核對=(27, 2, 2, 1, 18, 是, unit_mod_n_qualified);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-qualification · candidate
26. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":3,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 1]], 2, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
27. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":4,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 3], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 3], [2, 1]], 2, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
28. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":5,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 3], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 3], [4, 1]], 4, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
29. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":6,"a":5}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
30. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":7,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 1]], 3, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 1]], 3, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
31. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":8,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 3], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 3], [2, 1]], 2, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
32. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":9,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 7], [5, 5], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 7], [5, 5], [6, 1]], 6, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
33. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":10,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 3], [2, 9], [3, 7], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 3], [2, 9], [3, 7], [4, 1]], 4, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
34. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":11,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 5], [5, 10], [6, 9], [7, 7], [8, 3], [9, 6], [10, 1]], 10, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 5], [5, 10], [6, 9], [7, 7], [8, 3], [9, 6], [10, 1]], 10, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
35. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":12,"a":5}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
36. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":13,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 3], [5, 6], [6, 12], [7, 11], [8, 9], [9, 5], [10, 10], [11, 7], [12, 1]], 12, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 3], [5, 6], [6, 12], [7, 11], [8, 9], [9, 5], [10, 10], [11, 7], [12, 1]], 12, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
37. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":14,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 3], [2, 9], [3, 13], [4, 11], [5, 5], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 3], [2, 9], [3, 13], [4, 11], [5, 5], [6, 1]], 6, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
38. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":15,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 1]], 4, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
39. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":16,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 3], [2, 9], [3, 11], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 3], [2, 9], [3, 11], [4, 1]], 4, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
40. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":17,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 16], [5, 15], [6, 13], [7, 9], [8, 1]], 8, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 15], [6, 13], [7, 9], [8, 1]], 8, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
41. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":18,"a":5}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 5], [2, 7], [3, 17], [4, 13], [5, 11], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 5], [2, 7], [3, 17], [4, 13], [5, 11], [6, 1]], 6, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
42. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":19,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 16], [5, 13], [6, 7], [7, 14], [8, 9], [9, 18], [10, 17], [11, 15], [12, 11], [13, 3], [14, 6], [15, 12], [16, 5], [17, 10], [18, 1]], 18, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 13], [6, 7], [7, 14], [8, 9], [9, 18], [10, 17], [11, 15], [12, 11], [13, 3], [14, 6], [15, 12], [16, 5], [17, 10], [18, 1]], 18, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
43. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":20,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 3], [2, 9], [3, 7], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 3], [2, 9], [3, 7], [4, 1]], 4, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
44. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":21,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 16], [5, 11], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 11], [6, 1]], 6, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
45. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":22,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 3], [2, 9], [3, 5], [4, 15], [5, 1]], 5, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 3], [2, 9], [3, 5], [4, 15], [5, 1]], 5, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
46. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":23,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 16], [5, 9], [6, 18], [7, 13], [8, 3], [9, 6], [10, 12], [11, 1]], 11, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 9], [6, 18], [7, 13], [8, 3], [9, 6], [10, 12], [11, 1]], 11, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
47. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":24,"a":5}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
48. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":25,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 16], [5, 7], [6, 14], [7, 3], [8, 6], [9, 12], [10, 24], [11, 23], [12, 21], [13, 17], [14, 9], [15, 18], [16, 11], [17, 22], [18, 19], [19, 13], [20, 1]], 20, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 7], [6, 14], [7, 3], [8, 6], [9, 12], [10, 24], [11, 23], [12, 21], [13, 17], [14, 9], [15, 18], [16, 11], [17, 22], [18, 19], [19, 13], [20, 1]], 20, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
49. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":26,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 3], [2, 9], [3, 1]], 3, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 3], [2, 9], [3, 1]], 3, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
50. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return;資料={"n":27,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。
查看答案與解釋
([[1, 2], [2, 4], [3, 8], [4, 16], [5, 5], [6, 10], [7, 20], [8, 13], [9, 26], [10, 25], [11, 23], [12, 19], [13, 11], [14, 22], [15, 17], [16, 7], [17, 14], [18, 1]], 18, 是, 是, order_is_the_first_positive_return)
Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 5], [6, 10], [7, 20], [8, 13], [9, 26], [10, 25], [11, 23], [12, 19], [13, 11], [14, 22], [15, 17], [16, 7], [17, 14], [18, 1]], 18, 是, 是, order_is_the_first_positive_return);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-power-first-return · candidate
51. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":3,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(2, 2, 是, [[1, 2], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(2, 2, 是, [[1, 2], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
52. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":4,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(2, 2, 是, [[1, 3], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(2, 2, 是, [[1, 3], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
53. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":5,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(4, 4, 是, [[1, 2], [2, 4], [3, 3], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(4, 4, 是, [[1, 2], [2, 4], [3, 3], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
54. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":6,"a":5}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(2, 2, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(2, 2, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
55. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":7,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(3, 6, 是, [[1, 2], [2, 4], [3, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(3, 6, 是, [[1, 2], [2, 4], [3, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
56. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":8,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(2, 4, 是, [[1, 3], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(2, 4, 是, [[1, 3], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
57. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":9,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(6, 6, 是, [[1, 2], [2, 4], [3, 8], [4, 7], [5, 5], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(6, 6, 是, [[1, 2], [2, 4], [3, 8], [4, 7], [5, 5], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
58. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":10,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(4, 4, 是, [[1, 3], [2, 9], [3, 7], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(4, 4, 是, [[1, 3], [2, 9], [3, 7], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
59. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":11,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(10, 10, 是, [[1, 2], [2, 4], [3, 8], [4, 5], [5, 10], [6, 9], [7, 7], [8, 3], [9, 6], [10, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(10, 10, 是, [[1, 2], [2, 4], [3, 8], [4, 5], [5, 10], [6, 9], [7, 7], [8, 3], [9, 6], [10, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
60. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":12,"a":5}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(2, 4, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(2, 4, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
61. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":13,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(12, 12, 是, [[1, 2], [2, 4], [3, 8], [4, 3], [5, 6], [6, 12], [7, 11], [8, 9], [9, 5], [10, 10], [11, 7], [12, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(12, 12, 是, [[1, 2], [2, 4], [3, 8], [4, 3], [5, 6], [6, 12], [7, 11], [8, 9], [9, 5], [10, 10], [11, 7], [12, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
62. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":14,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(6, 6, 是, [[1, 3], [2, 9], [3, 13], [4, 11], [5, 5], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(6, 6, 是, [[1, 3], [2, 9], [3, 13], [4, 11], [5, 5], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
63. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":15,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(4, 8, 是, [[1, 2], [2, 4], [3, 8], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(4, 8, 是, [[1, 2], [2, 4], [3, 8], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
64. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":16,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(4, 8, 是, [[1, 3], [2, 9], [3, 11], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(4, 8, 是, [[1, 3], [2, 9], [3, 11], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
65. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":17,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(8, 16, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 15], [6, 13], [7, 9], [8, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(8, 16, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 15], [6, 13], [7, 9], [8, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
66. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":18,"a":5}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(6, 6, 是, [[1, 5], [2, 7], [3, 17], [4, 13], [5, 11], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(6, 6, 是, [[1, 5], [2, 7], [3, 17], [4, 13], [5, 11], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
67. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":19,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(18, 18, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 13], [6, 7], [7, 14], [8, 9], [9, 18], [10, 17], [11, 15], [12, 11], [13, 3], [14, 6], [15, 12], [16, 5], [17, 10], [18, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(18, 18, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 13], [6, 7], [7, 14], [8, 9], [9, 18], [10, 17], [11, 15], [12, 11], [13, 3], [14, 6], [15, 12], [16, 5], [17, 10], [18, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
68. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":20,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(4, 8, 是, [[1, 3], [2, 9], [3, 7], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(4, 8, 是, [[1, 3], [2, 9], [3, 7], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
69. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":21,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(6, 12, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 11], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(6, 12, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 11], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
70. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":22,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(5, 10, 是, [[1, 3], [2, 9], [3, 5], [4, 15], [5, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(5, 10, 是, [[1, 3], [2, 9], [3, 5], [4, 15], [5, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
71. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":23,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(11, 22, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 9], [6, 18], [7, 13], [8, 3], [9, 6], [10, 12], [11, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(11, 22, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 9], [6, 18], [7, 13], [8, 3], [9, 6], [10, 12], [11, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
72. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":24,"a":5}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(2, 8, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(2, 8, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
73. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":25,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(20, 20, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 7], [6, 14], [7, 3], [8, 6], [9, 12], [10, 24], [11, 23], [12, 21], [13, 17], [14, 9], [15, 18], [16, 11], [17, 22], [18, 19], [19, 13], [20, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(20, 20, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 7], [6, 14], [7, 3], [8, 6], [9, 12], [10, 24], [11, 23], [12, 21], [13, 17], [14, 9], [15, 18], [16, 11], [17, 22], [18, 19], [19, 13], [20, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
74. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":26,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(3, 12, 是, [[1, 3], [2, 9], [3, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(3, 12, 是, [[1, 3], [2, 9], [3, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
75. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility;資料={"n":27,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。
查看答案與解釋
(18, 18, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 5], [6, 10], [7, 20], [8, 13], [9, 26], [10, 25], [11, 23], [12, 19], [13, 11], [14, 22], [15, 17], [16, 7], [17, 14], [18, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)
Wave6第十層核對=(18, 18, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 5], [6, 10], [7, 20], [8, 13], [9, 26], [10, 25], [11, 23], [12, 19], [13, 11], [14, 22], [15, 17], [16, 7], [17, 14], [18, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-phi-divisibility · candidate
76. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"noncoprime","n":8,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(noncoprime, 8, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(noncoprime, 8, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
77. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"invalid_modulus","n":1,"a":3}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(invalid_modulus, 1, 3, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(invalid_modulus, 1, 3, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
78. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"order_one","n":5,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(order_one, 5, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(order_one, 5, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
79. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"negative_normalization","n":8,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(negative_normalization, 8, -1, 7, 1, 是, [[1, 7], [2, 1]], 2, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(negative_normalization, 8, -1, 7, 1, 是, [[1, 7], [2, 1]], 2, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
80. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"noncoprime","n":16,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(noncoprime, 16, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(noncoprime, 16, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
81. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"invalid_modulus","n":1,"a":7}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(invalid_modulus, 1, 7, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(invalid_modulus, 1, 7, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
82. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"order_one","n":9,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(order_one, 9, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(order_one, 9, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
83. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"negative_normalization","n":12,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(negative_normalization, 12, -1, 11, 1, 是, [[1, 11], [2, 1]], 2, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(negative_normalization, 12, -1, 11, 1, 是, [[1, 11], [2, 1]], 2, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
84. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"noncoprime","n":24,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(noncoprime, 24, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(noncoprime, 24, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
85. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"invalid_modulus","n":1,"a":11}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(invalid_modulus, 1, 11, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(invalid_modulus, 1, 11, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
86. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"order_one","n":13,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(order_one, 13, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(order_one, 13, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
87. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"negative_normalization","n":16,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(negative_normalization, 16, -1, 15, 1, 是, [[1, 15], [2, 1]], 2, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(negative_normalization, 16, -1, 15, 1, 是, [[1, 15], [2, 1]], 2, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
88. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"noncoprime","n":32,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(noncoprime, 32, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(noncoprime, 32, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
89. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"invalid_modulus","n":1,"a":15}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(invalid_modulus, 1, 15, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(invalid_modulus, 1, 15, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
90. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"order_one","n":17,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(order_one, 17, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(order_one, 17, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
91. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"negative_normalization","n":20,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(negative_normalization, 20, -1, 19, 1, 是, [[1, 19], [2, 1]], 2, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(negative_normalization, 20, -1, 19, 1, 是, [[1, 19], [2, 1]], 2, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
92. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"noncoprime","n":40,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(noncoprime, 40, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(noncoprime, 40, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
93. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"invalid_modulus","n":1,"a":19}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(invalid_modulus, 1, 19, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(invalid_modulus, 1, 19, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
94. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"order_one","n":21,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(order_one, 21, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(order_one, 21, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
95. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"negative_normalization","n":24,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(negative_normalization, 24, -1, 23, 1, 是, [[1, 23], [2, 1]], 2, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(negative_normalization, 24, -1, 23, 1, 是, [[1, 23], [2, 1]], 2, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
96. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"noncoprime","n":48,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(noncoprime, 48, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(noncoprime, 48, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
97. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"invalid_modulus","n":1,"a":23}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(invalid_modulus, 1, 23, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(invalid_modulus, 1, 23, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
98. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"order_one","n":25,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(order_one, 25, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(order_one, 25, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
99. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"negative_normalization","n":28,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(negative_normalization, 28, -1, 27, 1, 是, [[1, 27], [2, 1]], 2, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(negative_normalization, 28, -1, 27, 1, 是, [[1, 27], [2, 1]], 2, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
100. 模n乘法階
n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary;資料={"mode":"noncoprime","n":56,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。
invalid modulus以null記residue/gcd;noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null;order_one與negative則保留真正table。
查看答案與解釋
(noncoprime, 56, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)
Wave6第十層核對=(noncoprime, 56, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary);依完整明示資料獨立重算;complete power table與phi enumeration核order divides phi;noncoprime/invalid modulus undefined;不宣稱order=phi、primitive root或additive order。
本題逐步詳解(AI 二次覆核)
模n乘法階 · qf-mod47-morder-boundary · candidate
原始題目 JSONL · 驗收報告