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[组合] 求275的倍数个数

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lemondian Posted at ereyesterday 11:18 |Read mode
若将数字$1,2,3,5,6,7,8,9$排成一行,所得到的8位数中,是275的倍数的有多少个?

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realnumber Posted at ereyesterday 12:30
搜索一下被11,25整除的数的规律

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 Author| lemondian Posted at ereyesterday 14:19
realnumber 发表于 2025-4-17 12:30
搜索一下被11,25整除的数的规律
结果是多少?能否写个解答过程供学习一下?
谢谢

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l不想算,尾数为25或75, “奇数位和"与"偶素位和"之差为11的倍数,不会太多的  Posted at ereyesterday 18:44
144?  Posted at ereyesterday 21:05

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kuing Posted at ereyesterday 20:42

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kuing Posted at ereyesterday 22:25
(1)`abcdef25`
那就是要 `\{1,3,6,7,8,9\}=\{a,c,e\}\cup\{b,d,f\}` 且 `a+c+e+2-b-d-f-5=\pm11`。
不难列出只有 `\{7,8,9\}\cup\{1,3,6\}` 和 `\{1,3,9\}\cup\{6,7,8\}` 两种;

(2)`abcdef75`
那就是要 `\{1,2,3,6,8,9\}=\{a,c,e\}\cup\{b,d,f\}` 且 `a+c+e+7-b-d-f-5=\pm11`。
不难列出只有 `\{2,8,9\}\cup\{1,3,6\}` 一种。

综上,答案为 `3!\times3!\times3=108`。

mma 验证:
  1. Select[FromDigits /@ Permutations[{1, 2, 3, 5, 6, 7, 8, 9}], Mod[#, 275] == 0 &] // Length
Copy the Code

输出 108

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有没有奇数位的和与偶数位的和之差为0?  Posted at yesterday 10:03
6个数和是29,不可能等分吧  Posted at yesterday 10:39

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kuing Posted at yesterday 13:22
lemondian 点评
有没有奇数位的和与偶数位的和之差为0?

有没有认真看 4# 链接的解法啊,那个差的奇偶性一定是不变的啊

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2025-4-19 17:41 GMT+8

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