# 模n乘法階基礎 · 脈衝 R07

本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。

本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261015-advance-20261116-r07-multiplicative-order-mod-n-foundations · 2026-10-15 · 100 題

欄序為來源綁定的呈現補充；原始 JSONL 與數學答案不變。

[欄序與來源綁定資料](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":3,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(3, 2, 2, 1, 2, 是, unit_mod_n_qualified)

Wave6第十層核對=(3, 2, 2, 1, 2, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0001/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 2. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":4,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(4, 3, 3, 1, 2, 是, unit_mod_n_qualified)

Wave6第十層核對=(4, 3, 3, 1, 2, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0002/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 3. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":5,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(5, 2, 2, 1, 4, 是, unit_mod_n_qualified)

Wave6第十層核對=(5, 2, 2, 1, 4, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0003/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 4. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":6,"a":5}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(6, 5, 5, 1, 2, 是, unit_mod_n_qualified)

Wave6第十層核對=(6, 5, 5, 1, 2, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0004/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 5. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":7,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(7, 2, 2, 1, 6, 是, unit_mod_n_qualified)

Wave6第十層核對=(7, 2, 2, 1, 6, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0005/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 6. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":8,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(8, 3, 3, 1, 4, 是, unit_mod_n_qualified)

Wave6第十層核對=(8, 3, 3, 1, 4, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0006/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 7. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":9,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(9, 2, 2, 1, 6, 是, unit_mod_n_qualified)

Wave6第十層核對=(9, 2, 2, 1, 6, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0007/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 8. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":10,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(10, 3, 3, 1, 4, 是, unit_mod_n_qualified)

Wave6第十層核對=(10, 3, 3, 1, 4, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0008/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 9. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":11,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(11, 2, 2, 1, 10, 是, unit_mod_n_qualified)

Wave6第十層核對=(11, 2, 2, 1, 10, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0009/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 10. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":12,"a":5}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(12, 5, 5, 1, 4, 是, unit_mod_n_qualified)

Wave6第十層核對=(12, 5, 5, 1, 4, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0010/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 11. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":13,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(13, 2, 2, 1, 12, 是, unit_mod_n_qualified)

Wave6第十層核對=(13, 2, 2, 1, 12, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0011/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 12. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":14,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(14, 3, 3, 1, 6, 是, unit_mod_n_qualified)

Wave6第十層核對=(14, 3, 3, 1, 6, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0012/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 13. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":15,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(15, 2, 2, 1, 8, 是, unit_mod_n_qualified)

Wave6第十層核對=(15, 2, 2, 1, 8, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0013/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 14. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":16,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(16, 3, 3, 1, 8, 是, unit_mod_n_qualified)

Wave6第十層核對=(16, 3, 3, 1, 8, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0014/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 15. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":17,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(17, 2, 2, 1, 16, 是, unit_mod_n_qualified)

Wave6第十層核對=(17, 2, 2, 1, 16, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0015/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 16. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":18,"a":5}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(18, 5, 5, 1, 6, 是, unit_mod_n_qualified)

Wave6第十層核對=(18, 5, 5, 1, 6, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0016/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 17. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":19,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(19, 2, 2, 1, 18, 是, unit_mod_n_qualified)

Wave6第十層核對=(19, 2, 2, 1, 18, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0017/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 18. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":20,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(20, 3, 3, 1, 8, 是, unit_mod_n_qualified)

Wave6第十層核對=(20, 3, 3, 1, 8, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0018/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 19. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":21,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(21, 2, 2, 1, 12, 是, unit_mod_n_qualified)

Wave6第十層核對=(21, 2, 2, 1, 12, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0019/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 20. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":22,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(22, 3, 3, 1, 10, 是, unit_mod_n_qualified)

Wave6第十層核對=(22, 3, 3, 1, 10, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0020/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 21. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":23,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(23, 2, 2, 1, 22, 是, unit_mod_n_qualified)

Wave6第十層核對=(23, 2, 2, 1, 22, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0021/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 22. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":24,"a":5}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(24, 5, 5, 1, 8, 是, unit_mod_n_qualified)

Wave6第十層核對=(24, 5, 5, 1, 8, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0022/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 23. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":25,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(25, 2, 2, 1, 20, 是, unit_mod_n_qualified)

Wave6第十層核對=(25, 2, 2, 1, 20, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0023/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 24. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":26,"a":3}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(26, 3, 3, 1, 12, 是, unit_mod_n_qualified)

Wave6第十層核對=(26, 3, 3, 1, 12, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0024/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 25. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-qualification；資料={"n":27,"a":2}。請依 [n, a, normalized_residue, gcd, phi, unit_qualified, certificate] 欄序回答。

查看答案與解釋(27, 2, 2, 1, 18, 是, unit_mod_n_qualified)

Wave6第十層核對=(27, 2, 2, 1, 18, 是, unit_mod_n_qualified)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0025/)

模n乘法階 · qf-mod47-morder-qualification · candidate





## 26. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":3,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0026/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 27. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":4,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 3], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 3], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0027/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 28. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":5,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 3], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 3], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0028/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 29. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":6,"a":5}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0029/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 30. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":7,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 1]], 3, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 1]], 3, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0030/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 31. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":8,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 3], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 3], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0031/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 32. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":9,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 7], [5, 5], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 7], [5, 5], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0032/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 33. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":10,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 3], [2, 9], [3, 7], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 3], [2, 9], [3, 7], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0033/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 34. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":11,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 5], [5, 10], [6, 9], [7, 7], [8, 3], [9, 6], [10, 1]], 10, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 5], [5, 10], [6, 9], [7, 7], [8, 3], [9, 6], [10, 1]], 10, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0034/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 35. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":12,"a":5}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0035/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 36. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":13,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 3], [5, 6], [6, 12], [7, 11], [8, 9], [9, 5], [10, 10], [11, 7], [12, 1]], 12, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 3], [5, 6], [6, 12], [7, 11], [8, 9], [9, 5], [10, 10], [11, 7], [12, 1]], 12, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0036/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 37. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":14,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 3], [2, 9], [3, 13], [4, 11], [5, 5], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 3], [2, 9], [3, 13], [4, 11], [5, 5], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0037/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 38. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":15,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0038/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 39. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":16,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 3], [2, 9], [3, 11], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 3], [2, 9], [3, 11], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0039/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 40. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":17,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 16], [5, 15], [6, 13], [7, 9], [8, 1]], 8, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 15], [6, 13], [7, 9], [8, 1]], 8, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0040/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 41. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":18,"a":5}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 5], [2, 7], [3, 17], [4, 13], [5, 11], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 5], [2, 7], [3, 17], [4, 13], [5, 11], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0041/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 42. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":19,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 16], [5, 13], [6, 7], [7, 14], [8, 9], [9, 18], [10, 17], [11, 15], [12, 11], [13, 3], [14, 6], [15, 12], [16, 5], [17, 10], [18, 1]], 18, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 13], [6, 7], [7, 14], [8, 9], [9, 18], [10, 17], [11, 15], [12, 11], [13, 3], [14, 6], [15, 12], [16, 5], [17, 10], [18, 1]], 18, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0042/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 43. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":20,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 3], [2, 9], [3, 7], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 3], [2, 9], [3, 7], [4, 1]], 4, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0043/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 44. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":21,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 16], [5, 11], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 11], [6, 1]], 6, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0044/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 45. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":22,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 3], [2, 9], [3, 5], [4, 15], [5, 1]], 5, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 3], [2, 9], [3, 5], [4, 15], [5, 1]], 5, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0045/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 46. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":23,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 16], [5, 9], [6, 18], [7, 13], [8, 3], [9, 6], [10, 12], [11, 1]], 11, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 9], [6, 18], [7, 13], [8, 3], [9, 6], [10, 12], [11, 1]], 11, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0046/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 47. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":24,"a":5}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 5], [2, 1]], 2, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0047/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 48. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":25,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 16], [5, 7], [6, 14], [7, 3], [8, 6], [9, 12], [10, 24], [11, 23], [12, 21], [13, 17], [14, 9], [15, 18], [16, 11], [17, 22], [18, 19], [19, 13], [20, 1]], 20, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 7], [6, 14], [7, 3], [8, 6], [9, 12], [10, 24], [11, 23], [12, 21], [13, 17], [14, 9], [15, 18], [16, 11], [17, 22], [18, 19], [19, 13], [20, 1]], 20, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0048/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 49. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":26,"a":3}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 3], [2, 9], [3, 1]], 3, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 3], [2, 9], [3, 1]], 3, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0049/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 50. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-power-first-return；資料={"n":27,"a":2}。請依 [power_rows, order, no_earlier_return, returns_one, certificate] 欄序回答。

查看答案與解釋([[1, 2], [2, 4], [3, 8], [4, 16], [5, 5], [6, 10], [7, 20], [8, 13], [9, 26], [10, 25], [11, 23], [12, 19], [13, 11], [14, 22], [15, 17], [16, 7], [17, 14], [18, 1]], 18, 是, 是, order_is_the_first_positive_return)

Wave6第十層核對=([[1, 2], [2, 4], [3, 8], [4, 16], [5, 5], [6, 10], [7, 20], [8, 13], [9, 26], [10, 25], [11, 23], [12, 19], [13, 11], [14, 22], [15, 17], [16, 7], [17, 14], [18, 1]], 18, 是, 是, order_is_the_first_positive_return)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0050/)

模n乘法階 · qf-mod47-morder-power-first-return · candidate





## 51. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":3,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(2, 2, 是, [[1, 2], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(2, 2, 是, [[1, 2], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0051/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 52. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":4,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(2, 2, 是, [[1, 3], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(2, 2, 是, [[1, 3], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0052/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 53. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":5,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(4, 4, 是, [[1, 2], [2, 4], [3, 3], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(4, 4, 是, [[1, 2], [2, 4], [3, 3], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0053/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 54. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":6,"a":5}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(2, 2, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(2, 2, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0054/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 55. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":7,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(3, 6, 是, [[1, 2], [2, 4], [3, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(3, 6, 是, [[1, 2], [2, 4], [3, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0055/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 56. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":8,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(2, 4, 是, [[1, 3], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(2, 4, 是, [[1, 3], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0056/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 57. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":9,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(6, 6, 是, [[1, 2], [2, 4], [3, 8], [4, 7], [5, 5], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(6, 6, 是, [[1, 2], [2, 4], [3, 8], [4, 7], [5, 5], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0057/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 58. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":10,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(4, 4, 是, [[1, 3], [2, 9], [3, 7], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(4, 4, 是, [[1, 3], [2, 9], [3, 7], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0058/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 59. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":11,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(10, 10, 是, [[1, 2], [2, 4], [3, 8], [4, 5], [5, 10], [6, 9], [7, 7], [8, 3], [9, 6], [10, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(10, 10, 是, [[1, 2], [2, 4], [3, 8], [4, 5], [5, 10], [6, 9], [7, 7], [8, 3], [9, 6], [10, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0059/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 60. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":12,"a":5}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(2, 4, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(2, 4, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0060/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 61. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":13,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(12, 12, 是, [[1, 2], [2, 4], [3, 8], [4, 3], [5, 6], [6, 12], [7, 11], [8, 9], [9, 5], [10, 10], [11, 7], [12, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(12, 12, 是, [[1, 2], [2, 4], [3, 8], [4, 3], [5, 6], [6, 12], [7, 11], [8, 9], [9, 5], [10, 10], [11, 7], [12, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0061/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 62. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":14,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(6, 6, 是, [[1, 3], [2, 9], [3, 13], [4, 11], [5, 5], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(6, 6, 是, [[1, 3], [2, 9], [3, 13], [4, 11], [5, 5], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0062/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 63. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":15,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(4, 8, 是, [[1, 2], [2, 4], [3, 8], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(4, 8, 是, [[1, 2], [2, 4], [3, 8], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0063/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 64. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":16,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(4, 8, 是, [[1, 3], [2, 9], [3, 11], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(4, 8, 是, [[1, 3], [2, 9], [3, 11], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0064/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 65. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":17,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(8, 16, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 15], [6, 13], [7, 9], [8, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(8, 16, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 15], [6, 13], [7, 9], [8, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0065/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 66. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":18,"a":5}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(6, 6, 是, [[1, 5], [2, 7], [3, 17], [4, 13], [5, 11], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(6, 6, 是, [[1, 5], [2, 7], [3, 17], [4, 13], [5, 11], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0066/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 67. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":19,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(18, 18, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 13], [6, 7], [7, 14], [8, 9], [9, 18], [10, 17], [11, 15], [12, 11], [13, 3], [14, 6], [15, 12], [16, 5], [17, 10], [18, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(18, 18, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 13], [6, 7], [7, 14], [8, 9], [9, 18], [10, 17], [11, 15], [12, 11], [13, 3], [14, 6], [15, 12], [16, 5], [17, 10], [18, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0067/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 68. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":20,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(4, 8, 是, [[1, 3], [2, 9], [3, 7], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(4, 8, 是, [[1, 3], [2, 9], [3, 7], [4, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0068/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 69. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":21,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(6, 12, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 11], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(6, 12, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 11], [6, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0069/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 70. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":22,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(5, 10, 是, [[1, 3], [2, 9], [3, 5], [4, 15], [5, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(5, 10, 是, [[1, 3], [2, 9], [3, 5], [4, 15], [5, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0070/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 71. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":23,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(11, 22, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 9], [6, 18], [7, 13], [8, 3], [9, 6], [10, 12], [11, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(11, 22, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 9], [6, 18], [7, 13], [8, 3], [9, 6], [10, 12], [11, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0071/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 72. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":24,"a":5}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(2, 8, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(2, 8, 是, [[1, 5], [2, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0072/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 73. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":25,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(20, 20, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 7], [6, 14], [7, 3], [8, 6], [9, 12], [10, 24], [11, 23], [12, 21], [13, 17], [14, 9], [15, 18], [16, 11], [17, 22], [18, 19], [19, 13], [20, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(20, 20, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 7], [6, 14], [7, 3], [8, 6], [9, 12], [10, 24], [11, 23], [12, 21], [13, 17], [14, 9], [15, 18], [16, 11], [17, 22], [18, 19], [19, 13], [20, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0073/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 74. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":26,"a":3}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(3, 12, 是, [[1, 3], [2, 9], [3, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(3, 12, 是, [[1, 3], [2, 9], [3, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0074/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 75. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-phi-divisibility；資料={"n":27,"a":2}。請依 [order, phi, order_divides_phi, power_rows, certificate] 欄序回答。

查看答案與解釋(18, 18, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 5], [6, 10], [7, 20], [8, 13], [9, 26], [10, 25], [11, 23], [12, 19], [13, 11], [14, 22], [15, 17], [16, 7], [17, 14], [18, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)

Wave6第十層核對=(18, 18, 是, [[1, 2], [2, 4], [3, 8], [4, 16], [5, 5], [6, 10], [7, 20], [8, 13], [9, 26], [10, 25], [11, 23], [12, 19], [13, 11], [14, 22], [15, 17], [16, 7], [17, 14], [18, 1]], ord_n_a_divides_phi_n_without_claiming_equality_or_primitive_root)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0075/)

模n乘法階 · qf-mod47-morder-phi-divisibility · candidate





## 76. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"noncoprime","n":8,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(noncoprime, 8, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(noncoprime, 8, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0076/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 77. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"invalid_modulus","n":1,"a":3}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(invalid_modulus, 1, 3, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(invalid_modulus, 1, 3, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0077/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 78. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"order_one","n":5,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(order_one, 5, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(order_one, 5, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0078/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 79. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"negative_normalization","n":8,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(negative_normalization, 8, -1, 7, 1, 是, [[1, 7], [2, 1]], 2, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(negative_normalization, 8, -1, 7, 1, 是, [[1, 7], [2, 1]], 2, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0079/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 80. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"noncoprime","n":16,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(noncoprime, 16, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(noncoprime, 16, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0080/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 81. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"invalid_modulus","n":1,"a":7}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(invalid_modulus, 1, 7, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(invalid_modulus, 1, 7, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0081/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 82. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"order_one","n":9,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(order_one, 9, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(order_one, 9, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0082/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 83. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"negative_normalization","n":12,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(negative_normalization, 12, -1, 11, 1, 是, [[1, 11], [2, 1]], 2, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(negative_normalization, 12, -1, 11, 1, 是, [[1, 11], [2, 1]], 2, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0083/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 84. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"noncoprime","n":24,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(noncoprime, 24, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(noncoprime, 24, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0084/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 85. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"invalid_modulus","n":1,"a":11}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(invalid_modulus, 1, 11, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(invalid_modulus, 1, 11, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0085/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 86. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"order_one","n":13,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(order_one, 13, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(order_one, 13, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0086/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 87. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"negative_normalization","n":16,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(negative_normalization, 16, -1, 15, 1, 是, [[1, 15], [2, 1]], 2, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(negative_normalization, 16, -1, 15, 1, 是, [[1, 15], [2, 1]], 2, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0087/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 88. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"noncoprime","n":32,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(noncoprime, 32, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(noncoprime, 32, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0088/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 89. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"invalid_modulus","n":1,"a":15}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(invalid_modulus, 1, 15, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(invalid_modulus, 1, 15, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0089/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 90. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"order_one","n":17,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(order_one, 17, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(order_one, 17, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0090/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 91. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"negative_normalization","n":20,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(negative_normalization, 20, -1, 19, 1, 是, [[1, 19], [2, 1]], 2, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(negative_normalization, 20, -1, 19, 1, 是, [[1, 19], [2, 1]], 2, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0091/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 92. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"noncoprime","n":40,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(noncoprime, 40, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(noncoprime, 40, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0092/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 93. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"invalid_modulus","n":1,"a":19}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(invalid_modulus, 1, 19, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(invalid_modulus, 1, 19, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0093/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 94. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"order_one","n":21,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(order_one, 21, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(order_one, 21, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0094/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 95. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"negative_normalization","n":24,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(negative_normalization, 24, -1, 23, 1, 是, [[1, 23], [2, 1]], 2, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(negative_normalization, 24, -1, 23, 1, 是, [[1, 23], [2, 1]], 2, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0095/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 96. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"noncoprime","n":48,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(noncoprime, 48, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(noncoprime, 48, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0096/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 97. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"invalid_modulus","n":1,"a":23}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(invalid_modulus, 1, 23, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(invalid_modulus, 1, 23, null, null, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0097/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 98. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"order_one","n":25,"a":1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(order_one, 25, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(order_one, 25, 1, 1, 1, 是, [[1, 1]], 1, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0098/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 99. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"negative_normalization","n":28,"a":-1}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(negative_normalization, 28, -1, 27, 1, 是, [[1, 27], [2, 1]], 2, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(negative_normalization, 28, -1, 27, 1, 是, [[1, 27], [2, 1]], 2, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0099/)

模n乘法階 · qf-mod47-morder-boundary · candidate





## 100. 模n乘法階

n>=2、normalized coprime residue與first positive power return責任。操作=morder-boundary；資料={"mode":"noncoprime","n":56,"a":2}。請依 [mode, n, a, normalized_residue_or_null, gcd_or_null, unit_qualified, power_rows_or_null, order_or_undefined, boundary] 欄序回答。

invalid modulus以null記residue/gcd；noncoprime仍記normalized residue與gcd。兩者order皆undefined且power table為null；order_one與negative則保留真正table。

查看答案與解釋(noncoprime, 56, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)

Wave6第十層核對=(noncoprime, 56, 2, 2, 2, 否, null, undefined, mode_specific_multiplicative_order_boundary)；依完整明示資料獨立重算；complete power table與phi enumeration核order divides phi；noncoprime/invalid modulus undefined；不宣稱order=phi、primitive root或additive order。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r07-multiplicative-order-mod-n-0100/)

模n乘法階 · qf-mod47-morder-boundary · candidate



[原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261116-r07-multiplicative-order-mod-n-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261116-r07-multiplicative-order-mod-n-foundations/validation-report.json)