第 16635 題：Euler φ函數

原題與已知條件

正整數與完整1到n代表約定。操作=totient-coprime-residues；資料={"n":10}。請依 [n,coprime_residues,representative_convention] 欄序回答。

• G001 — 正整數與完整1到n代表約定。操作=totient-coprime-residues；資料={"n":10}。請依 [n,coprime_residues,representative_convention] 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-euler-totient-function

逐步推導

selected modulus representative1:original nonnegative Euclidean inputs=[1, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[1, 10, 0, 1], [10, 1, 10, 0]];terminal positive gcd=1

selected modulus:original representative k=1 in1..10;verified gcd(k,n)=1

coprime indicator=1;include representative=是

selected modulus representative2:original nonnegative Euclidean inputs=[2, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[2, 10, 0, 2], [10, 2, 5, 0]];terminal positive gcd=2

selected modulus:original representative k=2 in1..10;verified gcd(k,n)=2

coprime indicator=0;include representative=否

selected modulus representative3:original nonnegative Euclidean inputs=[3, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[3, 10, 0, 3], [10, 3, 3, 1], [3, 1, 3, 0]];terminal positive gcd=1

selected modulus:original representative k=3 in1..10;verified gcd(k,n)=1

coprime indicator=1;include representative=是

selected modulus representative4:original nonnegative Euclidean inputs=[4, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[4, 10, 0, 4], [10, 4, 2, 2], [4, 2, 2, 0]];terminal positive gcd=2

selected modulus:original representative k=4 in1..10;verified gcd(k,n)=2

coprime indicator=0;include representative=否

selected modulus representative5:original nonnegative Euclidean inputs=[5, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[5, 10, 0, 5], [10, 5, 2, 0]];terminal positive gcd=5

selected modulus:original representative k=5 in1..10;verified gcd(k,n)=5

coprime indicator=0;include representative=否

selected modulus representative6:original nonnegative Euclidean inputs=[6, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[6, 10, 0, 6], [10, 6, 1, 4], [6, 4, 1, 2], [4, 2, 2, 0]];terminal positive gcd=2

selected modulus:original representative k=6 in1..10;verified gcd(k,n)=2

coprime indicator=0;include representative=否

selected modulus representative7:original nonnegative Euclidean inputs=[7, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[7, 10, 0, 7], [10, 7, 1, 3], [7, 3, 2, 1], [3, 1, 3, 0]];terminal positive gcd=1

selected modulus:original representative k=7 in1..10;verified gcd(k,n)=1

coprime indicator=1;include representative=是

selected modulus representative8:original nonnegative Euclidean inputs=[8, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[8, 10, 0, 8], [10, 8, 1, 2], [8, 2, 4, 0]];terminal positive gcd=2

selected modulus:original representative k=8 in1..10;verified gcd(k,n)=2

coprime indicator=0;include representative=否

selected modulus representative9:original nonnegative Euclidean inputs=[9, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[9, 10, 0, 9], [10, 9, 1, 1], [9, 1, 9, 0]];terminal positive gcd=1

selected modulus:original representative k=9 in1..10;verified gcd(k,n)=1

coprime indicator=1;include representative=是

selected modulus representative10:original nonnegative Euclidean inputs=[10, 10]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[10, 10, 1, 0]];terminal positive gcd=10

selected modulus:original representative k=10 in1..10;verified gcd(k,n)=10

coprime indicator=0;include representative=否

selected modulus exact coprime indicator count := sum(1, 0, 1, 0, 0, 0, 1, 0, 1, 0)

selected modulus exact coprime indicator count = 4

selected modulus:complete representative range1..10;all selected coprime representatives=[1, 3, 7, 9]

phi(10)=4;representatives_1_through_n

逐步反向核對

• S022 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S021 · premise_and_result_audit — 以精確有理數重算 selected modulus exact coprime indicator count，不以小數近似替代。
• S020 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S019 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S018 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S017 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S016 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S015 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S014 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S013 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S012 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S011 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S010 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S009 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S008 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S007 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S006 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S005 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S004 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S003 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(10, [1, 3, 7, 9], representatives_1_through_n)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。