推 raymond92928: 謝謝，mod 9m那個a和一開始的那個a是不同的a嗎？ 12/07 20:16
→ Desperato : 啊對 標錯了 12/08 14:06
※ 引述《raymond92928 (raymond)》之銘言：
: https://i.imgur.com/IDp0LyW.jpg
: 主題是modular arithmetic
: 找不到嚴謹的證明方法
Let n = 2^a 5^b m, m not multiple of 2 or 5
Given a set S = {1, 10, 10^2, 10^3, ..., 10^(9m-1)}
|S| = m, every element x of S never a multiple of 9m
Thus x = a (mod 9m), a in T = {1, 2, 3, ..., 9m-1}, |T| = 9m-1
Therefore exists x != y in S, x = y (mod 9m)
Write x = 10^i, y = 10^j, we may assume i < j
Then (y-x) is a multiple of 9m
and 10^c (7/9) (y-x) is a multiple of n, c = max{a, b}
which is the desired number.
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