第 21372 題：整數唯一質因數分解

原題與已知條件

完整bounded integer factorization與exact primality。操作=upf-uniqueness-audit；資料={"n":474,"factors_left":[2,3,79],"factors_right":[79,2,3]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 完整bounded integer factorization與exact primality。操作=upf-uniqueness-audit；資料={"n":474,"factors_left":[2,3,79],"factors_right":[79,2,3]}。請依 exact_ordered_tuple_declared_by_interface 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-unique-prime-factorization

逐步推導

source operation=upf-uniqueness-audit;original snapshot and mathematics retained

explicit ordered source output fields=[left_valid, right_valid, sorted_left, sorted_right, divisibility_rows, uniqueness_certificate, certificate]

upf LEFT factor index0:ORIGINAL candidate factor=2;ALL possible trial divisors2..floor(sqrt(v))=[]

ALL [divisor,remainder] rows=[];prime=是

upf LEFT complete product at original factor0 := mul(1, 2)

upf LEFT complete product at original factor0 = 2

upf LEFT factor index1:ORIGINAL candidate factor=3;ALL possible trial divisors2..floor(sqrt(v))=[]

ALL [divisor,remainder] rows=[];prime=是

upf LEFT complete product at original factor1 := mul(2, 3)

upf LEFT complete product at original factor1 = 6

upf LEFT factor index2:ORIGINAL candidate factor=79;ALL possible trial divisors2..floor(sqrt(v))=[2, 3, 4, 5, 6, 7, 8]

ALL [divisor,remainder] rows=[[2, 1], [3, 1], [4, 3], [5, 4], [6, 1], [7, 2], [8, 7]];prime=是

upf LEFT complete product at original factor2 := mul(6, 79)

upf LEFT complete product at original factor2 = 474

upf LEFT:ORIGINAL factors=[2, 3, 79];ALL primality flags=[是, 是, 是];original n=474

empty product convention1;actual complete product=474;all_prime=是;valid_prime_product=是

upf RIGHT factor index0:ORIGINAL candidate factor=79;ALL possible trial divisors2..floor(sqrt(v))=[2, 3, 4, 5, 6, 7, 8]

ALL [divisor,remainder] rows=[[2, 1], [3, 1], [4, 3], [5, 4], [6, 1], [7, 2], [8, 7]];prime=是

upf RIGHT complete product at original factor0 := mul(1, 79)

upf RIGHT complete product at original factor0 = 79

upf RIGHT factor index1:ORIGINAL candidate factor=2;ALL possible trial divisors2..floor(sqrt(v))=[]

ALL [divisor,remainder] rows=[];prime=是

upf RIGHT complete product at original factor1 := mul(79, 2)

upf RIGHT complete product at original factor1 = 158

upf RIGHT factor index2:ORIGINAL candidate factor=3;ALL possible trial divisors2..floor(sqrt(v))=[]

ALL [divisor,remainder] rows=[];prime=是

upf RIGHT complete product at original factor2 := mul(158, 3)

upf RIGHT complete product at original factor2 = 474

upf RIGHT:ORIGINAL factors=[79, 2, 3];ALL primality flags=[是, 是, 是];original n=474

empty product convention1;actual complete product=474;all_prime=是;valid_prime_product=是

SORTED left occurrence index=0,factor=2;ORIGINAL right factors=[79, 2, 3];full right product=474

[p,p_divides_product,FIRST original right index,right factor,right factor equals p]=[2, 是, 1, 2, 是];same right index MAY be reused

SORTED left occurrence index=1,factor=3;ORIGINAL right factors=[79, 2, 3];full right product=474

[p,p_divides_product,FIRST original right index,right factor,right factor equals p]=[3, 是, 2, 3, 是];same right index MAY be reused

SORTED left occurrence index=2,factor=79;ORIGINAL right factors=[79, 2, 3];full right product=474

[p,p_divides_product,FIRST original right index,right factor,right factor equals p]=[79, 是, 0, 79, 是];same right index MAY be reused

ORIGINAL n=474;left valid=是,right valid=是;sorted multisets=[[2, 3, 79], [2, 3, 79]];local witness rows=[[2, 是, 1, 2, 是], [3, 是, 2, 3, 是], [79, 是, 0, 79, 是]]

joint uniqueness certificate=是;general proof:prime p divides other product;Euclid lemma gives equal prime factor;match ONE occurrence,cancel nonzero p,then induct on remaining factor count

逐步反向核對

• 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 — 以精確有理數重算 upf RIGHT complete product at original factor2，不以小數近似替代。
• 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 — 以精確有理數重算 upf RIGHT complete product at original factor1，不以小數近似替代。
• 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 — 以精確有理數重算 upf RIGHT complete product at original factor0，不以小數近似替代。
• 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 — 以精確有理數重算 upf LEFT complete product at original factor2，不以小數近似替代。
• 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 — 以精確有理數重算 upf LEFT complete product at original factor1，不以小數近似替代。
• 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 — 以精確有理數重算 upf LEFT complete product at original factor0，不以小數近似替代。
• 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

(是, 是, [2, 3, 79], [2, 3, 79], [[2, 是, 1, 2, 是], [3, 是, 2, 3, 是], [79, 是, 0, 79, 是]], 是, Euclid_lemma_witness_p_divides_product_implies_p_divides_a_prime_factor)

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。