第 16609 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-boundary-prime；資料={"n":9}。請依 [n,phi_n,classification] 欄序回答。

• G001 — 正整數與完整1到n代表約定。操作=totient-boundary-prime；資料={"n":9}。請依 [n,phi_n,classification] 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, 9]

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

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

coprime indicator=1;include representative=是

selected modulus representative2:original nonnegative Euclidean inputs=[2, 9]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[2, 9, 0, 2], [9, 2, 4, 1], [2, 1, 2, 0]];terminal positive gcd=1

selected modulus:original representative k=2 in1..9;verified gcd(k,n)=1

coprime indicator=1;include representative=是

selected modulus representative3:original nonnegative Euclidean inputs=[3, 9]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[3, 9, 0, 3], [9, 3, 3, 0]];terminal positive gcd=3

selected modulus:original representative k=3 in1..9;verified gcd(k,n)=3

coprime indicator=0;include representative=否

selected modulus representative4:original nonnegative Euclidean inputs=[4, 9]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[4, 9, 0, 4], [9, 4, 2, 1], [4, 1, 4, 0]];terminal positive gcd=1

selected modulus:original representative k=4 in1..9;verified gcd(k,n)=1

coprime indicator=1;include representative=是

selected modulus representative5:original nonnegative Euclidean inputs=[5, 9]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[5, 9, 0, 5], [9, 5, 1, 4], [5, 4, 1, 1], [4, 1, 4, 0]];terminal positive gcd=1

selected modulus:original representative k=5 in1..9;verified gcd(k,n)=1

coprime indicator=1;include representative=是

selected modulus representative6:original nonnegative Euclidean inputs=[6, 9]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[6, 9, 0, 6], [9, 6, 1, 3], [6, 3, 2, 0]];terminal positive gcd=3

selected modulus:original representative k=6 in1..9;verified gcd(k,n)=3

coprime indicator=0;include representative=否

selected modulus representative7:original nonnegative Euclidean inputs=[7, 9]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[7, 9, 0, 7], [9, 7, 1, 2], [7, 2, 3, 1], [2, 1, 2, 0]];terminal positive gcd=1

selected modulus:original representative k=7 in1..9;verified gcd(k,n)=1

coprime indicator=1;include representative=是

selected modulus representative8:original nonnegative Euclidean inputs=[8, 9]

division rows[a,b,q,r] with a=q*b+r and0<=r<b=[[8, 9, 0, 8], [9, 8, 1, 1], [8, 1, 8, 0]];terminal positive gcd=1

selected modulus:original representative k=8 in1..9;verified gcd(k,n)=1

coprime indicator=1;include representative=是

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

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

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

coprime indicator=0;include representative=否

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

selected modulus exact coprime indicator count = 6

selected modulus:complete representative range1..9;all selected coprime representatives=[1, 2, 4, 5, 7, 8]

phi(9)=6;representatives_1_through_n

original n=9;exact phi=6;first proper trial divisor up to floor(sqrt(n))=3

n=1 handled before primality;classification=composite_no_prime_shortcut

逐步反向核對

• S021 · 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。
• 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 — 以精確有理數重算 selected modulus exact coprime indicator count，不以小數近似替代。
• 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

(9, 6, composite_no_prime_shortcut)

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。