{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0001","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=0、b=5，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(0, 0)","explanation_zh":"歐式商與餘數=(0, 0)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132001,"parameters":{"a":0,"b":5,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[\"zero_dividend\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0002","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-7、b=3，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-3, 2)","explanation_zh":"歐式商與餘數=(-3, 2)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132002,"parameters":{"a":-7,"b":3,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-7,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0003","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=3、b=4，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(0, 3)","explanation_zh":"歐式商與餘數=(0, 3)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132003,"parameters":{"a":3,"b":4,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[3,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0004","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-4、b=5，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-1, 1)","explanation_zh":"歐式商與餘數=(-1, 1)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132004,"parameters":{"a":-4,"b":5,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-4,5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0005","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=5、b=6，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(0, 5)","explanation_zh":"歐式商與餘數=(0, 5)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132005,"parameters":{"a":5,"b":6,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[5,6]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0006","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-6、b=2，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-3, 0)","explanation_zh":"歐式商與餘數=(-3, 0)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132006,"parameters":{"a":-6,"b":2,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-6,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0007","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=7、b=3，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(2, 1)","explanation_zh":"歐式商與餘數=(2, 1)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132007,"parameters":{"a":7,"b":3,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[7,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0008","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-8、b=4，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-2, 0)","explanation_zh":"歐式商與餘數=(-2, 0)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132008,"parameters":{"a":-8,"b":4,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-8,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0009","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=9、b=5，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(1, 4)","explanation_zh":"歐式商與餘數=(1, 4)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132009,"parameters":{"a":9,"b":5,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[9,5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0010","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-10、b=6，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-2, 2)","explanation_zh":"歐式商與餘數=(-2, 2)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132010,"parameters":{"a":-10,"b":6,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-10,6]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0011","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=11、b=2，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(5, 1)","explanation_zh":"歐式商與餘數=(5, 1)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132011,"parameters":{"a":11,"b":2,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[11,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0012","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-12、b=3，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-4, 0)","explanation_zh":"歐式商與餘數=(-4, 0)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132012,"parameters":{"a":-12,"b":3,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-12,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0013","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=13、b=4，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(3, 1)","explanation_zh":"歐式商與餘數=(3, 1)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132013,"parameters":{"a":13,"b":4,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[13,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0014","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-14、b=5，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-3, 1)","explanation_zh":"歐式商與餘數=(-3, 1)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132014,"parameters":{"a":-14,"b":5,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-14,5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0015","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=15、b=6，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(2, 3)","explanation_zh":"歐式商與餘數=(2, 3)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132015,"parameters":{"a":15,"b":6,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[15,6]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0016","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-16、b=2，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-8, 0)","explanation_zh":"歐式商與餘數=(-8, 0)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132016,"parameters":{"a":-16,"b":2,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-16,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0017","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=17、b=3，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(5, 2)","explanation_zh":"歐式商與餘數=(5, 2)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132017,"parameters":{"a":17,"b":3,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[17,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0018","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-18、b=4，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-5, 2)","explanation_zh":"歐式商與餘數=(-5, 2)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132018,"parameters":{"a":-18,"b":4,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-18,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0019","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=19、b=5，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(3, 4)","explanation_zh":"歐式商與餘數=(3, 4)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132019,"parameters":{"a":19,"b":5,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[19,5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0020","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-20、b=6，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-4, 4)","explanation_zh":"歐式商與餘數=(-4, 4)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132020,"parameters":{"a":-20,"b":6,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-20,6]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0021","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=21、b=2，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(10, 1)","explanation_zh":"歐式商與餘數=(10, 1)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132021,"parameters":{"a":21,"b":2,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[21,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0022","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-22、b=3，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-8, 2)","explanation_zh":"歐式商與餘數=(-8, 2)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132022,"parameters":{"a":-22,"b":3,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-22,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0023","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=23、b=4，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(5, 3)","explanation_zh":"歐式商與餘數=(5, 3)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132023,"parameters":{"a":23,"b":4,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[23,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0024","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=-24、b=5，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(-5, 1)","explanation_zh":"歐式商與餘數=(-5, 1)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132024,"parameters":{"a":-24,"b":5,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[-24,5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0025","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"帶符號整數的歐式商餘","stem_zh":"有限整數協議；a=25、b=6，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。","answer":{"kind":"expression","value":"(4, 1)","explanation_zh":"歐式商與餘數=(4, 1)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["帶符號整數的歐式商餘"]},"variant":{"family_id":"qf-density13-integer-euclidean-divmod","seed":202610132025,"parameters":{"a":25,"b":6,"operation":"density13_integer_euclidean_divmod","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-euclidean-divmod/[25,6]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0026","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=0，有號步s1=0、s2=0。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(0, 0, 0)","explanation_zh":"終點淨移與步量=(0, 0, 0)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132026,"parameters":{"start":0,"s1":0,"s2":0,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[0,\"0\",\"0\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0027","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=1，有號步s1=3、s2=-3。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(1, 0, 6)","explanation_zh":"終點淨移與步量=(1, 0, 6)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132027,"parameters":{"start":1,"s1":3,"s2":-3,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[1,\"0\",\"6\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0028","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-10，有號步s1=3、s2=-3。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(-10, 0, 6)","explanation_zh":"終點淨移與步量=(-10, 0, 6)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132028,"parameters":{"start":-10,"s1":3,"s2":-3,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-10,\"0\",\"6\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0029","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-9，有號步s1=4、s2=-4。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(-9, 0, 8)","explanation_zh":"終點淨移與步量=(-9, 0, 8)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132029,"parameters":{"start":-9,"s1":4,"s2":-4,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-9,\"0\",\"8\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0030","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-8，有號步s1=5、s2=-5。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(-8, 0, 10)","explanation_zh":"終點淨移與步量=(-8, 0, 10)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132030,"parameters":{"start":-8,"s1":5,"s2":-5,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-8,\"0\",\"10\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0031","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-7，有號步s1=6、s2=-1。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(-2, 5, 7)","explanation_zh":"終點淨移與步量=(-2, 5, 7)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132031,"parameters":{"start":-7,"s1":6,"s2":-1,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-7,\"5\",\"7\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0032","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-6，有號步s1=7、s2=-2。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(-1, 5, 9)","explanation_zh":"終點淨移與步量=(-1, 5, 9)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132032,"parameters":{"start":-6,"s1":7,"s2":-2,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-6,\"5\",\"9\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0033","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-5，有號步s1=8、s2=-3。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(0, 5, 11)","explanation_zh":"終點淨移與步量=(0, 5, 11)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132033,"parameters":{"start":-5,"s1":8,"s2":-3,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-5,\"5\",\"11\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0034","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-4，有號步s1=9、s2=-4。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(1, 5, 13)","explanation_zh":"終點淨移與步量=(1, 5, 13)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132034,"parameters":{"start":-4,"s1":9,"s2":-4,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-4,\"5\",\"13\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0035","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-3，有號步s1=10、s2=-5。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(2, 5, 15)","explanation_zh":"終點淨移與步量=(2, 5, 15)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132035,"parameters":{"start":-3,"s1":10,"s2":-5,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-3,\"5\",\"15\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0036","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-2，有號步s1=11、s2=-1。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(8, 10, 12)","explanation_zh":"終點淨移與步量=(8, 10, 12)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132036,"parameters":{"start":-2,"s1":11,"s2":-1,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-2,\"10\",\"12\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0037","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=-1，有號步s1=12、s2=-2。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(9, 10, 14)","explanation_zh":"終點淨移與步量=(9, 10, 14)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132037,"parameters":{"start":-1,"s1":12,"s2":-2,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[-1,\"10\",\"14\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0038","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=0，有號步s1=13、s2=-3。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(10, 10, 16)","explanation_zh":"終點淨移與步量=(10, 10, 16)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132038,"parameters":{"start":0,"s1":13,"s2":-3,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[0,\"10\",\"16\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0039","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=1，有號步s1=14、s2=-4。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(11, 10, 18)","explanation_zh":"終點淨移與步量=(11, 10, 18)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132039,"parameters":{"start":1,"s1":14,"s2":-4,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[1,\"10\",\"18\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0040","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=2，有號步s1=15、s2=-5。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(12, 10, 20)","explanation_zh":"終點淨移與步量=(12, 10, 20)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132040,"parameters":{"start":2,"s1":15,"s2":-5,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[2,\"10\",\"20\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0041","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=3，有號步s1=16、s2=-1。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(18, 15, 17)","explanation_zh":"終點淨移與步量=(18, 15, 17)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132041,"parameters":{"start":3,"s1":16,"s2":-1,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[3,\"15\",\"17\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0042","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=4，有號步s1=17、s2=-2。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(19, 15, 19)","explanation_zh":"終點淨移與步量=(19, 15, 19)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132042,"parameters":{"start":4,"s1":17,"s2":-2,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[4,\"15\",\"19\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0043","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=5，有號步s1=18、s2=-3。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(20, 15, 21)","explanation_zh":"終點淨移與步量=(20, 15, 21)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132043,"parameters":{"start":5,"s1":18,"s2":-3,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[5,\"15\",\"21\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0044","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=6，有號步s1=19、s2=-4。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(21, 15, 23)","explanation_zh":"終點淨移與步量=(21, 15, 23)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132044,"parameters":{"start":6,"s1":19,"s2":-4,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[6,\"15\",\"23\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0045","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=7，有號步s1=20、s2=-5。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(22, 15, 25)","explanation_zh":"終點淨移與步量=(22, 15, 25)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132045,"parameters":{"start":7,"s1":20,"s2":-5,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[7,\"15\",\"25\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0046","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=8，有號步s1=21、s2=-1。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(28, 20, 22)","explanation_zh":"終點淨移與步量=(28, 20, 22)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132046,"parameters":{"start":8,"s1":21,"s2":-1,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[8,\"20\",\"22\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0047","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=9，有號步s1=22、s2=-2。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(29, 20, 24)","explanation_zh":"終點淨移與步量=(29, 20, 24)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132047,"parameters":{"start":9,"s1":22,"s2":-2,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[9,\"20\",\"24\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0048","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=10，有號步s1=23、s2=-3。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(30, 20, 26)","explanation_zh":"終點淨移與步量=(30, 20, 26)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132048,"parameters":{"start":10,"s1":23,"s2":-3,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[10,\"20\",\"26\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0049","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=11，有號步s1=24、s2=-4。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(31, 20, 28)","explanation_zh":"終點淨移與步量=(31, 20, 28)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132049,"parameters":{"start":11,"s1":24,"s2":-4,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[11,\"20\",\"28\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0050","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"兩步的終點、淨移與步量","stem_zh":"有限整數協議；start=12，有號步s1=25、s2=-5。依序求 (終點, 淨移s1+s2, |s1|+|s2|)。","answer":{"kind":"expression","value":"(32, 20, 30)","explanation_zh":"終點淨移與步量=(32, 20, 30)。終點累加有號步，步量另加絕對值；返回原點也可能有正步量，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["兩步的終點、淨移與步量"]},"variant":{"family_id":"qf-density13-integer-two-step-displacement","seed":202610132050,"parameters":{"start":12,"s1":25,"s2":-5,"operation":"density13_integer_two_step_displacement","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-two-step-displacement/[12,\"20\",\"30\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0051","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=0、b=0。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(0, 0, 0, 0)","explanation_zh":"相反數與雙向差=(0, 0, 0, 0)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132051,"parameters":{"a":0,"b":0,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[0,0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0052","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-11、b=0。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(11, 0, -11, 11)","explanation_zh":"相反數與雙向差=(11, 0, -11, 11)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132052,"parameters":{"a":-11,"b":0,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-11,0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0053","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-10、b=-1。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(10, 1, -9, 9)","explanation_zh":"相反數與雙向差=(10, 1, -9, 9)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132053,"parameters":{"a":-10,"b":-1,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-10,-1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0054","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-9、b=-2。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(9, 2, -7, 7)","explanation_zh":"相反數與雙向差=(9, 2, -7, 7)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132054,"parameters":{"a":-9,"b":-2,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-9,-2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0055","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-8、b=-3。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(8, 3, -5, 5)","explanation_zh":"相反數與雙向差=(8, 3, -5, 5)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132055,"parameters":{"a":-8,"b":-3,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-8,-3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0056","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-7、b=-4。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(7, 4, -3, 3)","explanation_zh":"相反數與雙向差=(7, 4, -3, 3)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132056,"parameters":{"a":-7,"b":-4,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-7,-4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0057","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-6、b=-5。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(6, 5, -1, 1)","explanation_zh":"相反數與雙向差=(6, 5, -1, 1)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132057,"parameters":{"a":-6,"b":-5,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-6,-5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0058","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-5、b=-6。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(5, 6, 1, -1)","explanation_zh":"相反數與雙向差=(5, 6, 1, -1)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132058,"parameters":{"a":-5,"b":-6,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-5,-6]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0059","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-4、b=-7。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(4, 7, 3, -3)","explanation_zh":"相反數與雙向差=(4, 7, 3, -3)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132059,"parameters":{"a":-4,"b":-7,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-4,-7]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0060","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-3、b=-8。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(3, 8, 5, -5)","explanation_zh":"相反數與雙向差=(3, 8, 5, -5)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132060,"parameters":{"a":-3,"b":-8,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-3,-8]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0061","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-2、b=-9。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(2, 9, 7, -7)","explanation_zh":"相反數與雙向差=(2, 9, 7, -7)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132061,"parameters":{"a":-2,"b":-9,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-2,-9]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0062","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=-1、b=-10。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(1, 10, 9, -9)","explanation_zh":"相反數與雙向差=(1, 10, 9, -9)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132062,"parameters":{"a":-1,"b":-10,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[-1,-10]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0063","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=0、b=-11。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(0, 11, 11, -11)","explanation_zh":"相反數與雙向差=(0, 11, 11, -11)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132063,"parameters":{"a":0,"b":-11,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[0,-11]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0064","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=1、b=-12。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-1, 12, 13, -13)","explanation_zh":"相反數與雙向差=(-1, 12, 13, -13)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132064,"parameters":{"a":1,"b":-12,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[1,-12]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0065","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=2、b=-13。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-2, 13, 15, -15)","explanation_zh":"相反數與雙向差=(-2, 13, 15, -15)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132065,"parameters":{"a":2,"b":-13,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[2,-13]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0066","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=3、b=-14。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-3, 14, 17, -17)","explanation_zh":"相反數與雙向差=(-3, 14, 17, -17)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132066,"parameters":{"a":3,"b":-14,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[3,-14]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0067","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=4、b=-15。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-4, 15, 19, -19)","explanation_zh":"相反數與雙向差=(-4, 15, 19, -19)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132067,"parameters":{"a":4,"b":-15,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[4,-15]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0068","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=5、b=-16。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-5, 16, 21, -21)","explanation_zh":"相反數與雙向差=(-5, 16, 21, -21)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132068,"parameters":{"a":5,"b":-16,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[5,-16]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0069","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=6、b=-17。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-6, 17, 23, -23)","explanation_zh":"相反數與雙向差=(-6, 17, 23, -23)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132069,"parameters":{"a":6,"b":-17,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[6,-17]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0070","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=7、b=-18。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-7, 18, 25, -25)","explanation_zh":"相反數與雙向差=(-7, 18, 25, -25)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132070,"parameters":{"a":7,"b":-18,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[7,-18]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0071","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=8、b=-19。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-8, 19, 27, -27)","explanation_zh":"相反數與雙向差=(-8, 19, 27, -27)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132071,"parameters":{"a":8,"b":-19,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[8,-19]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0072","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=9、b=-20。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-9, 20, 29, -29)","explanation_zh":"相反數與雙向差=(-9, 20, 29, -29)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132072,"parameters":{"a":9,"b":-20,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[9,-20]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0073","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=10、b=-21。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-10, 21, 31, -31)","explanation_zh":"相反數與雙向差=(-10, 21, 31, -31)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132073,"parameters":{"a":10,"b":-21,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[10,-21]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0074","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=11、b=-22。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-11, 22, 33, -33)","explanation_zh":"相反數與雙向差=(-11, 22, 33, -33)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132074,"parameters":{"a":11,"b":-22,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[11,-22]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0075","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"相反數與兩個方向的差","stem_zh":"有限整數協議；a=12、b=-23。依序求 (−a, −b, a−b, b−a)，保留方向與符號。","answer":{"kind":"expression","value":"(-12, 23, 35, -35)","explanation_zh":"相反數與雙向差=(-12, 23, 35, -35)。相反數逐項換號，差按指定方向計算；有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["相反數與兩個方向的差"]},"variant":{"family_id":"qf-density13-integer-opposite-difference","seed":202610132075,"parameters":{"a":12,"b":-23,"operation":"density13_integer_opposite_difference","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-opposite-difference/[12,-23]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0076","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"四個有理候選的整數資格與分組","stem_zh":"有限整數協議；原序fractions=[0/1, -2/1, 1/2, 1/-3]，q≠0。依序列 (整數資格是／否列表, 整數實值列表, 非整數既約有理值列表)，分組仍保留原序。","answer":{"kind":"expression","value":"([是, 是, 否, 否], [0, -2], [1/2, -1/3])","explanation_zh":"整數資格與分組值=([是, 是, 否, 否], [0, -2], [1/2, -1/3])。先規約至正分母，分母一才是整數；負分母也可表示有效值，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["四個有理候選的整數資格與分組"]},"variant":{"family_id":"qf-density13-integer-rational-grid-split","seed":202610132076,"parameters":{"fractions":[[0,1],[-2,1],[1,2],[1,-3]],"operation":"density13_integer_rational_grid_split","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-rational-grid-split/[\"0/1\",\"-2/1\",\"1/2\",\"-1/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0077","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"四個有理候選的整數資格與分組","stem_zh":"有限整數協議；原序fractions=[2/-2, 3/1, 4/-2, 5/2]，q≠0。依序列 (整數資格是／否列表, 整數實值列表, 非整數既約有理值列表)，分組仍保留原序。","answer":{"kind":"expression","value":"([是, 是, 是, 否], [-1, 3, -2], [5/2])","explanation_zh":"整數資格與分組值=([是, 是, 是, 否], [-1, 3, -2], [5/2])。先規約至正分母，分母一才是整數；負分母也可表示有效值，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["四個有理候選的整數資格與分組"]},"variant":{"family_id":"qf-density13-integer-rational-grid-split","seed":202610132077,"parameters":{"fractions":[[2,-2],[3,1],[4,-2],[5,2]],"operation":"density13_integer_rational_grid_split","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-rational-grid-split/[\"-1/1\",\"3/1\",\"-2/1\",\"5/2\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0078","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"四個有理候選的整數資格與分組","stem_zh":"有限整數協議；原序fractions=[-10/1, 4/-2, 3/7, -4/-9]，q≠0。依序列 (整數資格是／否列表, 整數實值列表, 非整數既約有理值列表)，分組仍保留原序。","answer":{"kind":"expression","value":"([是, 是, 否, 否], [-10, -2], [3/7, 4/9])","explanation_zh":"整數資格與分組值=([是, 是, 否, 否], [-10, -2], [3/7, 4/9])。先規約至正分母，分母一才是整數；負分母也可表示有效值，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["四個有理候選的整數資格與分組"]},"variant":{"family_id":"qf-density13-integer-rational-grid-split","seed":202610132078,"parameters":{"fractions":[[-10,1],[4,-2],[3,7],[-4,-9]],"operation":"density13_integer_rational_grid_split","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-rational-grid-split/[\"-10/1\",\"-2/1\",\"3/7\",\"4/9\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0079","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"四個有理候選的整數資格與分組","stem_zh":"有限整數協議；原序fractions=[-9/1, 6/-2, 4/9, -5/-11]，q≠0。依序列 (整數資格是／否列表, 整數實值列表, 非整數既約有理值列表)，分組仍保留原序。","answer":{"kind":"expression","value":"([是, 是, 否, 否], [-9, -3], [4/9, 5/11])","explanation_zh":"整數資格與分組值=([是, 是, 否, 否], [-9, -3], [4/9, 5/11])。先規約至正分母，分母一才是整數；負分母也可表示有效值，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["四個有理候選的整數資格與分組"]},"variant":{"family_id":"qf-density13-integer-rational-grid-split","seed":202610132079,"parameters":{"fractions":[[-9,1],[6,-2],[4,9],[-5,-11]],"operation":"density13_integer_rational_grid_split","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-rational-grid-split/[\"-9/1\",\"-3/1\",\"4/9\",\"5/11\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0080","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"四個有理候選的整數資格與分組","stem_zh":"有限整數協議；原序fractions=[-8/1, 8/-2, 5/11, -6/-13]，q≠0。依序列 (整數資格是／否列表, 整數實值列表, 非整數既約有理值列表)，分組仍保留原序。","answer":{"kind":"expression","value":"([是, 是, 否, 否], [-8, -4], [5/11, 6/13])","explanation_zh":"整數資格與分組值=([是, 是, 否, 否], [-8, -4], [5/11, 6/13])。先規約至正分母，分母一才是整數；負分母也可表示有效值，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["四個有理候選的整數資格與分組"]},"variant":{"family_id":"qf-density13-integer-rational-grid-split","seed":202610132080,"parameters":{"fractions":[[-8,1],[8,-2],[5,11],[-6,-13]],"operation":"density13_integer_rational_grid_split","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-rational-grid-split/[\"-8/1\",\"-4/1\",\"5/11\",\"6/13\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0081","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"四個有理候選的整數資格與分組","stem_zh":"有限整數協議；原序fractions=[-7/1, 10/-2, 6/13, -7/-15]，q≠0。依序列 (整數資格是／否列表, 整數實值列表, 非整數既約有理值列表)，分組仍保留原序。","answer":{"kind":"expression","value":"([是, 是, 否, 否], [-7, -5], [6/13, 7/15])","explanation_zh":"整數資格與分組值=([是, 是, 否, 否], [-7, -5], [6/13, 7/15])。先規約至正分母，分母一才是整數；負分母也可表示有效值，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["四個有理候選的整數資格與分組"]},"variant":{"family_id":"qf-density13-integer-rational-grid-split","seed":202610132081,"parameters":{"fractions":[[-7,1],[10,-2],[6,13],[-7,-15]],"operation":"density13_integer_rational_grid_split","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-rational-grid-split/[\"-7/1\",\"-5/1\",\"6/13\",\"7/15\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0082","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"四個有理候選的整數資格與分組","stem_zh":"有限整數協議；原序fractions=[-6/1, 12/-2, 7/15, -8/-17]，q≠0。依序列 (整數資格是／否列表, 整數實值列表, 非整數既約有理值列表)，分組仍保留原序。","answer":{"kind":"expression","value":"([是, 是, 否, 否], [-6, -6], [7/15, 8/17])","explanation_zh":"整數資格與分組值=([是, 是, 否, 否], [-6, -6], [7/15, 8/17])。先規約至正分母，分母一才是整數；負分母也可表示有效值，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["四個有理候選的整數資格與分組"]},"variant":{"family_id":"qf-density13-integer-rational-grid-split","seed":202610132082,"parameters":{"fractions":[[-6,1],[12,-2],[7,15],[-8,-17]],"operation":"density13_integer_rational_grid_split","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-rational-grid-split/[\"-6/1\",\"-6/1\",\"7/15\",\"8/17\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0083","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"四個有理候選的整數資格與分組","stem_zh":"有限整數協議；原序fractions=[-5/1, 14/-2, 8/17, -9/-19]，q≠0。依序列 (整數資格是／否列表, 整數實值列表, 非整數既約有理值列表)，分組仍保留原序。","answer":{"kind":"expression","value":"([是, 是, 否, 否], [-5, -7], [8/17, 9/19])","explanation_zh":"整數資格與分組值=([是, 是, 否, 否], [-5, -7], [8/17, 9/19])。先規約至正分母，分母一才是整數；負分母也可表示有效值，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["四個有理候選的整數資格與分組"]},"variant":{"family_id":"qf-density13-integer-rational-grid-split","seed":202610132083,"parameters":{"fractions":[[-5,1],[14,-2],[8,17],[-9,-19]],"operation":"density13_integer_rational_grid_split","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-rational-grid-split/[\"-5/1\",\"-7/1\",\"8/17\",\"9/19\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r02-integer-0084","batch_id":"batch-20261009-advance-20261013-r02-integer-foundations","language":"zh-Hant","source_mko_ids":["mko-integer"],"title_zh":"四個有理候選的整數資格與分組","stem_zh":"有限整數協議；原序fractions=[-4/1, 16/-2, 9/19, -10/-21]，q≠0。依序列 (整數資格是／否列表, 整數實值列表, 非整數既約有理值列表)，分組仍保留原序。","answer":{"kind":"expression","value":"([是, 是, 否, 否], [-4, -8], [9/19, 10/21])","explanation_zh":"整數資格與分組值=([是, 是, 否, 否], [-4, -8], [9/19, 10/21])。先規約至正分母，分母一才是整數；負分母也可表示有效值，有限運算非整數環公理證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["四個有理候選的整數資格與分組"]},"variant":{"family_id":"qf-density13-integer-rational-grid-split","seed":202610132084,"parameters":{"fractions":[[-4,1],[16,-2],[9,19],[-10,-21]],"operation":"density13_integer_rational_grid_split","domain":"finite_integer_arithmetic_protocol","semantic_template":"qf-density13-integer-rational-grid-split/[\"-4/1\",\"-8/1\",\"9/19\",\"10/21\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
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