# 有限含一環基礎 · 脈衝 R01

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-20261014-advance-20261027-r01-finite-unital-ring-foundations · 2026-10-14 · 100 題



## 1. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,4,1,2,4],[0,3,0,3,0,3],[0,4,2,0,4,2],[0,5,4,3,2,1]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)

Wave4根節點核對=(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0001/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 2. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,3,5,0,4],[2,3,5,4,1,0],[3,5,4,0,2,1],[4,0,1,2,5,3],[5,4,0,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,5,0,5,2],[0,3,0,3,3,0],[0,4,5,3,1,2],[0,5,2,0,2,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0002/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 3. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,4,5,3,0],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,3,5,1,0,2],[5,0,1,4,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,3,2,0,3],[0,3,2,3,0,2],[0,4,0,0,4,4],[0,5,3,2,4,1]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0003/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 4. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,4,0,5,3],[2,4,5,1,3,0],[3,0,1,5,2,4],[4,5,3,2,0,1],[5,3,0,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,5,5,0,2],[0,3,5,1,4,2],[0,4,0,4,4,0],[0,5,2,2,0,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0004/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 5. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,5,4,0,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,0,1,5,3,2],[5,3,4,1,2,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,3,2,3,0],[0,3,2,3,2,0],[0,4,3,2,1,5],[0,5,0,0,5,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0005/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 6. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,5,0,3,4],[2,5,4,1,0,3],[3,0,1,4,5,2],[4,3,0,5,2,1],[5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,4,5,2,0],[0,3,4,1,2,5],[0,4,2,2,4,0],[0,5,0,5,0,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)

Wave4根節點核對=(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0006/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 7. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,4,2,5,0],[2,4,0,5,1,3],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,0,3,1,2,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,0,2],[0,3,0,4,3,4],[0,4,0,3,4,3],[0,5,2,4,3,1]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0007/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 8. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,5,2,0,4],[2,5,0,4,3,1],[3,2,4,5,1,0],[4,0,3,1,5,2],[5,4,1,0,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,2,0],[0,3,0,5,5,3],[0,4,2,5,1,3],[0,5,0,3,3,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0008/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 9. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,5,4,2,0],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,2,1,5,0,3],[5,0,4,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,3,0,3],[0,3,3,2,0,2],[0,4,0,0,4,4],[0,5,3,2,4,1]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0009/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 10. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,0,4,5,2],[2,0,5,1,3,4],[3,4,1,5,2,0],[4,5,3,2,0,1],[5,2,4,0,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,5,4,3],[0,3,5,5,0,3],[0,4,4,0,4,0],[0,5,3,3,0,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0010/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 11. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,4,5,0,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,0,5,1,2,3],[5,2,1,4,3,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,4,3,0],[0,3,3,2,2,0],[0,4,3,2,1,5],[0,5,0,0,5,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)

Wave4根節點核對=(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0011/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 12. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,0,5,2,4],[2,0,4,1,5,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,4,3,5],[0,3,4,4,3,0],[0,4,3,3,4,0],[0,5,5,0,0,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0012/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 13. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,5,2,0],[2,3,0,1,5,4],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,0,4,2,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,0,2],[0,3,0,3,4,4],[0,4,0,4,3,3],[0,5,2,4,3,1]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0013/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 14. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,5,0,2,3],[2,5,0,4,3,1],[3,0,4,5,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,2,0,0],[0,3,2,1,5,4],[0,4,0,5,5,4],[0,5,0,4,4,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0014/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 15. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,5,2,3,0],[2,5,4,1,0,3],[3,2,1,0,5,4],[4,3,0,5,2,1],[5,0,3,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,4,4],[0,3,0,3,0,3],[0,4,4,0,2,2],[0,5,4,3,2,1]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0015/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 16. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,0,5,3,2],[2,0,5,4,1,3],[3,5,4,0,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,4,5,4],[0,3,3,3,0,0],[0,4,5,0,5,4],[0,5,4,0,4,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)

Wave4根節點核對=(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0016/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 17. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,0,5,2],[2,3,4,5,0,1],[3,0,5,2,1,4],[4,5,0,1,2,3],[5,2,1,4,3,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,4,4,0],[0,3,4,1,2,5],[0,4,4,2,2,0],[0,5,0,5,0,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0017/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 18. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,0,2,5,3],[2,0,3,5,1,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,3,4,1,2,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,4,3,5],[0,3,4,3,4,0],[0,4,3,4,3,0],[0,5,5,0,0,5]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0018/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 19. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,3,4,0,2],[2,3,0,1,5,4],[3,4,1,5,2,0],[4,0,5,2,3,1],[5,2,4,0,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,2,0],[0,3,0,3,5,5],[0,4,2,5,1,3],[0,5,0,5,3,3]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0019/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 20. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,4,0,3,2],[2,4,0,5,1,3],[3,0,5,4,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,2,0,0],[0,3,2,1,5,4],[0,4,0,5,4,5],[0,5,0,4,5,4]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0020/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 21. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,4,2,0,3],[2,4,5,1,3,0],[3,2,1,0,5,4],[4,0,3,5,2,1],[5,3,0,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,1,5,5],[0,3,0,3,3,0],[0,4,5,3,1,2],[0,5,5,0,2,2]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)

Wave4根節點核對=(6, 0, 1, 是, 否, 否, 否, 是, 否, multiplication_associativity)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0021/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 22. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,0,4,2,3],[2,0,4,5,3,1],[3,4,5,0,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,3,5,4],[0,3,3,3,0,0],[0,4,5,0,4,5],[0,5,4,0,5,4]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0022/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 23. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,3,0,2,4],[2,3,5,4,1,0],[3,0,4,2,5,1],[4,2,1,5,0,3],[5,4,0,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,5,0,5],[0,3,5,1,4,2],[0,4,0,4,4,0],[0,5,5,2,0,2]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0023/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 24. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,0,2,3,4],[2,0,3,4,5,1],[3,2,4,5,1,0],[4,3,5,1,0,2],[5,4,1,0,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,5,4,3],[0,3,5,3,0,5],[0,4,4,0,4,0],[0,5,3,5,0,3]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 1, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 1, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0024/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 25. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-qualification；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,5,0,2],[2,3,1,4,5,0],[3,5,4,0,2,1],[4,0,5,2,1,3],[5,2,0,1,3,4]],"multiplication":[[0,0,0,0,0,0],[0,4,1,0,1,4],[0,1,2,3,4,5],[0,0,3,3,0,3],[0,1,4,0,4,1],[0,4,5,3,1,2]]}。請依 [n,zero,one,additive_abelian,multiplication_associative,left_distributive,right_distributive,distinct_identities,qualified,first_failure] 欄序回答。

查看答案與解釋(6, 0, 2, 是, 是, 是, 是, 是, 是, null)

Wave4根節點核對=(6, 0, 2, 是, 是, 是, 是, 是, 是, null)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0025/)

有限含一環 · qf-mod27-ring-qualification · candidate





## 26. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,4,0,2,4],[0,3,0,3,0,3],[0,4,2,0,4,2],[0,5,4,3,2,1]],"element":1}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(1, 0, 5, 0, 0)

Wave4根節點核對=(1, 0, 5, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0026/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 27. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,3,5,0,4],[2,3,5,4,1,0],[3,5,4,0,2,1],[4,0,1,2,5,3],[5,4,0,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,5,0,5,2],[0,3,0,3,3,0],[0,4,5,3,1,2],[0,5,2,0,2,5]],"element":0}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(0, 0, 0, 0, 0)

Wave4根節點核對=(0, 0, 0, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0027/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 28. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,4,5,3,0],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,3,5,1,0,2],[5,0,1,4,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,3,2,0,3],[0,3,2,3,0,2],[0,4,0,0,4,4],[0,5,3,2,4,1]],"element":5}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(5, 0, 1, 0, 0)

Wave4根節點核對=(5, 0, 1, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0028/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 29. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,4,0,5,3],[2,4,5,1,3,0],[3,0,1,5,2,4],[4,5,3,2,0,1],[5,3,0,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,5,5,0,2],[0,3,5,1,4,2],[0,4,0,4,4,0],[0,5,2,2,0,5]],"element":4}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(4, 0, 4, 0, 0)

Wave4根節點核對=(4, 0, 4, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0029/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 30. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,5,4,0,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,0,1,5,3,2],[5,3,4,1,2,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,3,2,3,0],[0,3,2,3,2,0],[0,4,3,2,1,5],[0,5,0,0,5,5]],"element":3}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(3, 0, 2, 0, 0)

Wave4根節點核對=(3, 0, 2, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0030/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 31. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,5,0,3,4],[2,5,4,1,0,3],[3,0,1,4,5,2],[4,3,0,5,2,1],[5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,4,4,2,0],[0,3,4,1,2,5],[0,4,2,2,4,0],[0,5,0,5,0,5]],"element":2}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(2, 0, 4, 0, 0)

Wave4根節點核對=(2, 0, 4, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0031/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 32. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,4,2,5,0],[2,4,0,5,1,3],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,0,3,1,2,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,0,2],[0,3,0,4,3,4],[0,4,0,3,4,3],[0,5,2,4,3,1]],"element":1}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(1, 0, 5, 0, 0)

Wave4根節點核對=(1, 0, 5, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0032/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 33. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,5,2,0,4],[2,5,0,4,3,1],[3,2,4,5,1,0],[4,0,3,1,5,2],[5,4,1,0,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,2,0],[0,3,0,5,5,3],[0,4,2,5,1,3],[0,5,0,3,3,5]],"element":0}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(0, 0, 0, 0, 0)

Wave4根節點核對=(0, 0, 0, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0033/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 34. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,5,4,2,0],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,2,1,5,0,3],[5,0,4,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,3,0,3],[0,3,3,2,0,2],[0,4,0,0,4,4],[0,5,3,2,4,1]],"element":5}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(5, 0, 1, 0, 0)

Wave4根節點核對=(5, 0, 1, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0034/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 35. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,0,4,5,2],[2,0,5,1,3,4],[3,4,1,5,2,0],[4,5,3,2,0,1],[5,2,4,0,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,5,4,3],[0,3,5,5,0,3],[0,4,4,0,4,0],[0,5,3,3,0,5]],"element":4}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(4, 0, 4, 0, 0)

Wave4根節點核對=(4, 0, 4, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0035/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 36. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,4,5,0,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,0,5,1,2,3],[5,2,1,4,3,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,3,3,0],[0,3,3,2,2,0],[0,4,3,2,1,5],[0,5,0,0,5,5]],"element":3}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(3, 0, 2, 0, 0)

Wave4根節點核對=(3, 0, 2, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0036/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 37. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,0,5,2,4],[2,0,4,1,5,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,4,3,5],[0,3,4,4,3,0],[0,4,3,3,4,0],[0,5,5,0,0,5]],"element":2}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(2, 0, 1, 0, 0)

Wave4根節點核對=(2, 0, 1, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0037/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 38. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,5,2,0],[2,3,0,1,5,4],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,0,4,2,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,0,2],[0,3,0,3,4,4],[0,4,0,4,3,3],[0,5,2,4,3,1]],"element":1}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(1, 0, 5, 0, 0)

Wave4根節點核對=(1, 0, 5, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0038/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 39. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,5,0,2,3],[2,5,0,4,3,1],[3,0,4,5,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,2,0,0],[0,3,2,1,5,4],[0,4,0,5,5,4],[0,5,0,4,4,5]],"element":0}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(0, 0, 0, 0, 0)

Wave4根節點核對=(0, 0, 0, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0039/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 40. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,5,2,3,0],[2,5,4,1,0,3],[3,2,1,0,5,4],[4,3,0,5,2,1],[5,0,3,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,4,4],[0,3,0,3,0,3],[0,4,4,0,2,2],[0,5,4,3,2,1]],"element":5}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(5, 0, 1, 0, 0)

Wave4根節點核對=(5, 0, 1, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0040/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 41. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,0,5,3,2],[2,0,5,4,1,3],[3,5,4,0,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,3,5,4],[0,3,3,3,0,0],[0,4,5,0,5,4],[0,5,4,0,4,5]],"element":4}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(4, 0, 5, 0, 0)

Wave4根節點核對=(4, 0, 5, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0041/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 42. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,0,5,2],[2,3,4,5,0,1],[3,0,5,2,1,4],[4,5,0,1,2,3],[5,2,1,4,3,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,4,4,0],[0,3,4,1,2,5],[0,4,4,2,2,0],[0,5,0,5,0,5]],"element":3}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(3, 0, 1, 0, 0)

Wave4根節點核對=(3, 0, 1, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0042/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 43. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,0,2,5,3],[2,0,3,5,1,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,3,4,1,2,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,4,3,5],[0,3,4,3,4,0],[0,4,3,4,3,0],[0,5,5,0,0,5]],"element":2}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(2, 0, 1, 0, 0)

Wave4根節點核對=(2, 0, 1, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0043/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 44. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,3,4,0,2],[2,3,0,1,5,4],[3,4,1,5,2,0],[4,0,5,2,3,1],[5,2,4,0,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,2,0],[0,3,0,3,5,5],[0,4,2,5,1,3],[0,5,0,5,3,3]],"element":1}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(1, 0, 4, 0, 0)

Wave4根節點核對=(1, 0, 4, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0044/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 45. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,4,0,3,2],[2,4,0,5,1,3],[3,0,5,4,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,2,0,0],[0,3,2,1,5,4],[0,4,0,5,4,5],[0,5,0,4,5,4]],"element":0}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(0, 0, 0, 0, 0)

Wave4根節點核對=(0, 0, 0, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0045/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 46. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,4,2,0,3],[2,4,5,1,3,0],[3,2,1,0,5,4],[4,0,3,5,2,1],[5,3,0,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,5,5],[0,3,0,3,3,0],[0,4,5,3,1,2],[0,5,5,0,2,2]],"element":5}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(5, 0, 2, 0, 0)

Wave4根節點核對=(5, 0, 2, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0046/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 47. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,0,4,2,3],[2,0,4,5,3,1],[3,4,5,0,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,3,5,4],[0,3,3,3,0,0],[0,4,5,0,4,5],[0,5,4,0,5,4]],"element":4}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(4, 0, 5, 0, 0)

Wave4根節點核對=(4, 0, 5, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0047/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 48. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,3,0,2,4],[2,3,5,4,1,0],[3,0,4,2,5,1],[4,2,1,5,0,3],[5,4,0,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,5,0,5],[0,3,5,1,4,2],[0,4,0,4,4,0],[0,5,5,2,0,2]],"element":3}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(3, 0, 1, 0, 0)

Wave4根節點核對=(3, 0, 1, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0048/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 49. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,0,2,3,4],[2,0,3,4,5,1],[3,2,4,5,1,0],[4,3,5,1,0,2],[5,4,1,0,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,5,4,3],[0,3,5,3,0,5],[0,4,4,0,4,0],[0,5,3,5,0,3]],"element":2}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(2, 0, 1, 0, 0)

Wave4根節點核對=(2, 0, 1, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0049/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 50. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-additive-inverse；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,5,0,2],[2,3,1,4,5,0],[3,5,4,0,2,1],[4,0,5,2,1,3],[5,2,0,1,3,4]],"multiplication":[[0,0,0,0,0,0],[0,4,1,0,1,4],[0,1,2,3,4,5],[0,0,3,3,0,3],[0,1,4,0,4,1],[0,4,5,3,1,2]],"element":1}。請依 [element,zero,inverse,left_sum,right_sum] 欄序回答。

查看答案與解釋(1, 0, 4, 0, 0)

Wave4根節點核對=(1, 0, 4, 0, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0050/)

有限含一環 · qf-mod27-ring-additive-inverse · candidate





## 51. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,4,0,2,4],[0,3,0,3,0,3],[0,4,2,0,4,2],[0,5,4,3,2,1]],"element":1,"exponent":1}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(1, 1, 1, 1)

Wave4根節點核對=(1, 1, 1, 1)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0051/)

有限含一環 · qf-mod27-ring-power · candidate





## 52. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,3,5,0,4],[2,3,5,4,1,0],[3,5,4,0,2,1],[4,0,1,2,5,3],[5,4,0,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,5,0,5,2],[0,3,0,3,3,0],[0,4,5,3,1,2],[0,5,2,0,2,5]],"element":0,"exponent":2}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(0, 2, 1, 0)

Wave4根節點核對=(0, 2, 1, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0052/)

有限含一環 · qf-mod27-ring-power · candidate





## 53. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,4,5,3,0],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,3,5,1,0,2],[5,0,1,4,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,3,2,0,3],[0,3,2,3,0,2],[0,4,0,0,4,4],[0,5,3,2,4,1]],"element":5,"exponent":3}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(5, 3, 1, 5)

Wave4根節點核對=(5, 3, 1, 5)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0053/)

有限含一環 · qf-mod27-ring-power · candidate





## 54. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,4,0,5,3],[2,4,5,1,3,0],[3,0,1,5,2,4],[4,5,3,2,0,1],[5,3,0,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,5,5,0,2],[0,3,5,1,4,2],[0,4,0,4,4,0],[0,5,2,2,0,5]],"element":4,"exponent":4}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(4, 4, 1, 4)

Wave4根節點核對=(4, 4, 1, 4)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0054/)

有限含一環 · qf-mod27-ring-power · candidate





## 55. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,5,4,0,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,0,1,5,3,2],[5,3,4,1,2,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,3,2,3,0],[0,3,2,3,2,0],[0,4,3,2,1,5],[0,5,0,0,5,5]],"element":3,"exponent":5}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(3, 5, 1, 3)

Wave4根節點核對=(3, 5, 1, 3)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0055/)

有限含一環 · qf-mod27-ring-power · candidate





## 56. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,5,0,3,4],[2,5,4,1,0,3],[3,0,1,4,5,2],[4,3,0,5,2,1],[5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,4,4,2,0],[0,3,4,1,2,5],[0,4,2,2,4,0],[0,5,0,5,0,5]],"element":2,"exponent":6}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(2, 6, 1, 4)

Wave4根節點核對=(2, 6, 1, 4)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0056/)

有限含一環 · qf-mod27-ring-power · candidate





## 57. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,4,2,5,0],[2,4,0,5,1,3],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,0,3,1,2,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,0,2],[0,3,0,4,3,4],[0,4,0,3,4,3],[0,5,2,4,3,1]],"element":1,"exponent":7}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(1, 7, 1, 1)

Wave4根節點核對=(1, 7, 1, 1)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0057/)

有限含一環 · qf-mod27-ring-power · candidate





## 58. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,5,2,0,4],[2,5,0,4,3,1],[3,2,4,5,1,0],[4,0,3,1,5,2],[5,4,1,0,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,2,0],[0,3,0,5,5,3],[0,4,2,5,1,3],[0,5,0,3,3,5]],"element":0,"exponent":8}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(0, 8, 1, 0)

Wave4根節點核對=(0, 8, 1, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0058/)

有限含一環 · qf-mod27-ring-power · candidate





## 59. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,5,4,2,0],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,2,1,5,0,3],[5,0,4,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,3,0,3],[0,3,3,2,0,2],[0,4,0,0,4,4],[0,5,3,2,4,1]],"element":5,"exponent":9}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(5, 9, 1, 5)

Wave4根節點核對=(5, 9, 1, 5)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0059/)

有限含一環 · qf-mod27-ring-power · candidate





## 60. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,0,4,5,2],[2,0,5,1,3,4],[3,4,1,5,2,0],[4,5,3,2,0,1],[5,2,4,0,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,5,4,3],[0,3,5,5,0,3],[0,4,4,0,4,0],[0,5,3,3,0,5]],"element":4,"exponent":10}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(4, 10, 1, 4)

Wave4根節點核對=(4, 10, 1, 4)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0060/)

有限含一環 · qf-mod27-ring-power · candidate





## 61. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,4,5,0,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,0,5,1,2,3],[5,2,1,4,3,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,3,3,0],[0,3,3,2,2,0],[0,4,3,2,1,5],[0,5,0,0,5,5]],"element":3,"exponent":11}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(3, 11, 1, 3)

Wave4根節點核對=(3, 11, 1, 3)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0061/)

有限含一環 · qf-mod27-ring-power · candidate





## 62. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,0,5,2,4],[2,0,4,1,5,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,4,3,5],[0,3,4,4,3,0],[0,4,3,3,4,0],[0,5,5,0,0,5]],"element":2,"exponent":12}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(2, 12, 1, 1)

Wave4根節點核對=(2, 12, 1, 1)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0062/)

有限含一環 · qf-mod27-ring-power · candidate





## 63. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,5,2,0],[2,3,0,1,5,4],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,0,4,2,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,0,2],[0,3,0,3,4,4],[0,4,0,4,3,3],[0,5,2,4,3,1]],"element":1,"exponent":13}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(1, 13, 1, 1)

Wave4根節點核對=(1, 13, 1, 1)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0063/)

有限含一環 · qf-mod27-ring-power · candidate





## 64. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,5,0,2,3],[2,5,0,4,3,1],[3,0,4,5,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,2,0,0],[0,3,2,1,5,4],[0,4,0,5,5,4],[0,5,0,4,4,5]],"element":0,"exponent":14}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(0, 14, 1, 0)

Wave4根節點核對=(0, 14, 1, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0064/)

有限含一環 · qf-mod27-ring-power · candidate





## 65. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,5,2,3,0],[2,5,4,1,0,3],[3,2,1,0,5,4],[4,3,0,5,2,1],[5,0,3,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,4,4],[0,3,0,3,0,3],[0,4,4,0,2,2],[0,5,4,3,2,1]],"element":5,"exponent":15}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(5, 15, 1, 5)

Wave4根節點核對=(5, 15, 1, 5)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0065/)

有限含一環 · qf-mod27-ring-power · candidate





## 66. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,0,5,3,2],[2,0,5,4,1,3],[3,5,4,0,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,3,5,4],[0,3,3,3,0,0],[0,4,5,0,5,4],[0,5,4,0,4,5]],"element":4,"exponent":16}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(4, 16, 1, 5)

Wave4根節點核對=(4, 16, 1, 5)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0066/)

有限含一環 · qf-mod27-ring-power · candidate





## 67. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,0,5,2],[2,3,4,5,0,1],[3,0,5,2,1,4],[4,5,0,1,2,3],[5,2,1,4,3,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,4,4,0],[0,3,4,1,2,5],[0,4,4,2,2,0],[0,5,0,5,0,5]],"element":3,"exponent":17}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(3, 17, 1, 3)

Wave4根節點核對=(3, 17, 1, 3)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0067/)

有限含一環 · qf-mod27-ring-power · candidate





## 68. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,0,2,5,3],[2,0,3,5,1,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,3,4,1,2,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,4,3,5],[0,3,4,3,4,0],[0,4,3,4,3,0],[0,5,5,0,0,5]],"element":2,"exponent":18}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(2, 18, 1, 1)

Wave4根節點核對=(2, 18, 1, 1)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0068/)

有限含一環 · qf-mod27-ring-power · candidate





## 69. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,3,4,0,2],[2,3,0,1,5,4],[3,4,1,5,2,0],[4,0,5,2,3,1],[5,2,4,0,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,2,0],[0,3,0,3,5,5],[0,4,2,5,1,3],[0,5,0,5,3,3]],"element":1,"exponent":19}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(1, 19, 1, 1)

Wave4根節點核對=(1, 19, 1, 1)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0069/)

有限含一環 · qf-mod27-ring-power · candidate





## 70. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,4,0,3,2],[2,4,0,5,1,3],[3,0,5,4,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,2,0,0],[0,3,2,1,5,4],[0,4,0,5,4,5],[0,5,0,4,5,4]],"element":0,"exponent":20}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(0, 20, 1, 0)

Wave4根節點核對=(0, 20, 1, 0)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0070/)

有限含一環 · qf-mod27-ring-power · candidate





## 71. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,4,2,0,3],[2,4,5,1,3,0],[3,2,1,0,5,4],[4,0,3,5,2,1],[5,3,0,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,5,5],[0,3,0,3,3,0],[0,4,5,3,1,2],[0,5,5,0,2,2]],"element":5,"exponent":21}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(5, 21, 1, 5)

Wave4根節點核對=(5, 21, 1, 5)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0071/)

有限含一環 · qf-mod27-ring-power · candidate





## 72. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,0,4,2,3],[2,0,4,5,3,1],[3,4,5,0,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,3,5,4],[0,3,3,3,0,0],[0,4,5,0,4,5],[0,5,4,0,5,4]],"element":4,"exponent":22}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(4, 22, 1, 4)

Wave4根節點核對=(4, 22, 1, 4)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0072/)

有限含一環 · qf-mod27-ring-power · candidate





## 73. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,3,0,2,4],[2,3,5,4,1,0],[3,0,4,2,5,1],[4,2,1,5,0,3],[5,4,0,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,5,0,5],[0,3,5,1,4,2],[0,4,0,4,4,0],[0,5,5,2,0,2]],"element":3,"exponent":23}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(3, 23, 1, 3)

Wave4根節點核對=(3, 23, 1, 3)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0073/)

有限含一環 · qf-mod27-ring-power · candidate





## 74. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,0,2,3,4],[2,0,3,4,5,1],[3,2,4,5,1,0],[4,3,5,1,0,2],[5,4,1,0,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,5,4,3],[0,3,5,3,0,5],[0,4,4,0,4,0],[0,5,3,5,0,3]],"element":2,"exponent":24}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(2, 24, 1, 1)

Wave4根節點核對=(2, 24, 1, 1)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0074/)

有限含一環 · qf-mod27-ring-power · candidate





## 75. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-power；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,5,0,2],[2,3,1,4,5,0],[3,5,4,0,2,1],[4,0,5,2,1,3],[5,2,0,1,3,4]],"multiplication":[[0,0,0,0,0,0],[0,4,1,0,1,4],[0,1,2,3,4,5],[0,0,3,3,0,3],[0,1,4,0,4,1],[0,4,5,3,1,2]],"element":1,"exponent":25}。請依 [element,exponent,one,result] 欄序回答。

查看答案與解釋(1, 25, 2, 1)

Wave4根節點核對=(1, 25, 2, 1)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0075/)

有限含一環 · qf-mod27-ring-power · candidate





## 76. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,3,4,5,0],[2,3,4,5,0,1],[3,4,5,0,1,2],[4,5,0,1,2,3],[5,0,1,2,3,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,4,0,2,4],[0,3,0,3,0,3],[0,4,2,0,4,2],[0,5,4,3,2,1]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0076/)

有限含一環 · qf-mod27-ring-identities · candidate





## 77. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,3,5,0,4],[2,3,5,4,1,0],[3,5,4,0,2,1],[4,0,1,2,5,3],[5,4,0,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,5,0,5,2],[0,3,0,3,3,0],[0,4,5,3,1,2],[0,5,2,0,2,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0077/)

有限含一環 · qf-mod27-ring-identities · candidate





## 78. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,4,5,3,0],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,3,5,1,0,2],[5,0,1,4,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,3,2,0,3],[0,3,2,3,0,2],[0,4,0,0,4,4],[0,5,3,2,4,1]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0078/)

有限含一環 · qf-mod27-ring-identities · candidate





## 79. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,4,0,5,3],[2,4,5,1,3,0],[3,0,1,5,2,4],[4,5,3,2,0,1],[5,3,0,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,5,5,0,2],[0,3,5,1,4,2],[0,4,0,4,4,0],[0,5,2,2,0,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0079/)

有限含一環 · qf-mod27-ring-identities · candidate





## 80. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,5,4,0,3],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,0,1,5,3,2],[5,3,4,1,2,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,3,2,3,0],[0,3,2,3,2,0],[0,4,3,2,1,5],[0,5,0,0,5,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0080/)

有限含一環 · qf-mod27-ring-identities · candidate





## 81. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,2,5,0,3,4],[2,5,4,1,0,3],[3,0,1,4,5,2],[4,3,0,5,2,1],[5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,4,4,2,0],[0,3,4,1,2,5],[0,4,2,2,4,0],[0,5,0,5,0,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0081/)

有限含一環 · qf-mod27-ring-identities · candidate





## 82. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,4,2,5,0],[2,4,0,5,1,3],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,0,3,1,2,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,0,2],[0,3,0,4,3,4],[0,4,0,3,4,3],[0,5,2,4,3,1]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0082/)

有限含一環 · qf-mod27-ring-identities · candidate





## 83. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,5,2,0,4],[2,5,0,4,3,1],[3,2,4,5,1,0],[4,0,3,1,5,2],[5,4,1,0,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,2,0],[0,3,0,5,5,3],[0,4,2,5,1,3],[0,5,0,3,3,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0083/)

有限含一環 · qf-mod27-ring-identities · candidate





## 84. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,5,4,2,0],[2,5,3,0,1,4],[3,4,0,2,5,1],[4,2,1,5,0,3],[5,0,4,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,3,0,3],[0,3,3,2,0,2],[0,4,0,0,4,4],[0,5,3,2,4,1]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0084/)

有限含一環 · qf-mod27-ring-identities · candidate





## 85. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,0,4,5,2],[2,0,5,1,3,4],[3,4,1,5,2,0],[4,5,3,2,0,1],[5,2,4,0,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,5,4,3],[0,3,5,5,0,3],[0,4,4,0,4,0],[0,5,3,3,0,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0085/)

有限含一環 · qf-mod27-ring-identities · candidate





## 86. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,4,5,0,2],[2,4,3,0,5,1],[3,5,0,2,1,4],[4,0,5,1,2,3],[5,2,1,4,3,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,3,3,0],[0,3,3,2,2,0],[0,4,3,2,1,5],[0,5,0,0,5,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0086/)

有限含一環 · qf-mod27-ring-identities · candidate





## 87. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,3,0,5,2,4],[2,0,4,1,5,3],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,4,3,5],[0,3,4,4,3,0],[0,4,3,3,4,0],[0,5,5,0,0,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0087/)

有限含一環 · qf-mod27-ring-identities · candidate





## 88. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,5,2,0],[2,3,0,1,5,4],[3,5,1,4,0,2],[4,2,5,0,3,1],[5,0,4,2,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,0,2],[0,3,0,3,4,4],[0,4,0,4,3,3],[0,5,2,4,3,1]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0088/)

有限含一環 · qf-mod27-ring-identities · candidate





## 89. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,5,0,2,3],[2,5,0,4,3,1],[3,0,4,5,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,2,0,0],[0,3,2,1,5,4],[0,4,0,5,5,4],[0,5,0,4,4,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0089/)

有限含一環 · qf-mod27-ring-identities · candidate





## 90. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,5,2,3,0],[2,5,4,1,0,3],[3,2,1,0,5,4],[4,3,0,5,2,1],[5,0,3,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,4,4],[0,3,0,3,0,3],[0,4,4,0,2,2],[0,5,4,3,2,1]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0090/)

有限含一環 · qf-mod27-ring-identities · candidate





## 91. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,0,5,3,2],[2,0,5,4,1,3],[3,5,4,0,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,3,5,4],[0,3,3,3,0,0],[0,4,5,0,5,4],[0,5,4,0,4,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0091/)

有限含一環 · qf-mod27-ring-identities · candidate





## 92. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,0,5,2],[2,3,4,5,0,1],[3,0,5,2,1,4],[4,5,0,1,2,3],[5,2,1,4,3,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,4,4,0],[0,3,4,1,2,5],[0,4,4,2,2,0],[0,5,0,5,0,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0092/)

有限含一環 · qf-mod27-ring-identities · candidate





## 93. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,0,2,5,3],[2,0,3,5,1,4],[3,2,5,4,0,1],[4,5,1,0,3,2],[5,3,4,1,2,0]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,4,3,5],[0,3,4,3,4,0],[0,4,3,4,3,0],[0,5,5,0,0,5]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0093/)

有限含一環 · qf-mod27-ring-identities · candidate





## 94. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,3,4,0,2],[2,3,0,1,5,4],[3,4,1,5,2,0],[4,0,5,2,3,1],[5,2,4,0,1,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,2,0],[0,3,0,3,5,5],[0,4,2,5,1,3],[0,5,0,5,3,3]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0094/)

有限含一環 · qf-mod27-ring-identities · candidate





## 95. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,4,0,3,2],[2,4,0,5,1,3],[3,0,5,4,2,1],[4,3,1,2,5,0],[5,2,3,1,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,2,0,0],[0,3,2,1,5,4],[0,4,0,5,4,5],[0,5,0,4,5,4]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0095/)

有限含一環 · qf-mod27-ring-identities · candidate





## 96. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,4,2,0,3],[2,4,5,1,3,0],[3,2,1,0,5,4],[4,0,3,5,2,1],[5,3,0,4,1,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,0,5,5],[0,3,0,3,3,0],[0,4,5,3,1,2],[0,5,5,0,2,2]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0096/)

有限含一環 · qf-mod27-ring-identities · candidate





## 97. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,0,4,2,3],[2,0,4,5,3,1],[3,4,5,0,1,2],[4,2,3,1,5,0],[5,3,1,2,0,4]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,3,5,4],[0,3,3,3,0,0],[0,4,5,0,4,5],[0,5,4,0,5,4]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0097/)

有限含一環 · qf-mod27-ring-identities · candidate





## 98. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,3,0,2,4],[2,3,5,4,1,0],[3,0,4,2,5,1],[4,2,1,5,0,3],[5,4,0,1,3,2]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,2,5,0,5],[0,3,5,1,4,2],[0,4,0,4,4,0],[0,5,5,2,0,2]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0098/)

有限含一環 · qf-mod27-ring-identities · candidate





## 99. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,5,0,2,3,4],[2,0,3,4,5,1],[3,2,4,5,1,0],[4,3,5,1,0,2],[5,4,1,0,2,3]],"multiplication":[[0,0,0,0,0,0],[0,1,2,3,4,5],[0,2,1,5,4,3],[0,3,5,3,0,5],[0,4,4,0,4,0],[0,5,3,5,0,3]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 1, 是)

Wave4根節點核對=(0, 1, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0099/)

有限含一環 · qf-mod27-ring-identities · candidate





## 100. 有限含一環

同一完整有限載體上的加法表與乘法表。操作=ring-identities；資料={"n":6,"addition":[[0,1,2,3,4,5],[1,4,3,5,0,2],[2,3,1,4,5,0],[3,5,4,0,2,1],[4,0,5,2,1,3],[5,2,0,1,3,4]],"multiplication":[[0,0,0,0,0,0],[0,4,1,0,1,4],[0,1,2,3,4,5],[0,0,3,3,0,3],[0,1,4,0,4,1],[0,4,5,3,1,2]]}。請依 [zero,one,distinct] 欄序回答。

查看答案與解釋(0, 2, 是)

Wave4根節點核對=(0, 2, 是)；依完整明示資料重算；完整有限／精確座標切片不自動升格為任意無限定理；全部環公理逐對或逐三元組核對；含一環不自動是交換環整環或域。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261027-r01-finite-unital-ring-0100/)

有限含一環 · qf-mod27-ring-identities · candidate



[原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261027-r01-finite-unital-ring-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261027-r01-finite-unital-ring-foundations/validation-report.json)