判断能否被 9 整除

一个整数能被 9 整除，当且仅当其十进制各位数字之和能被 9 整除。

因为 $10\equiv 1\pmod 9$，所以

$$a_n10^n+\cdots+a_110+a_0\equiv a_n+\cdots+a_1+a_0\pmod 9.$$

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Original Explanation

An integer is divisible by 9 if and only if the sum of its digits is divisible by 9. Formally, if the integer is written in decimal as $$a_n 10^n + a_{n-1} 10^{n-1} + \cdots + a_1 10 + a_0,$$ then it is divisible by 9 exactly when $a_n + a_{n-1} + \cdots + a_1 + a_0$ is divisible by 9.