Competition problem — Number theory

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Number theory Difficulty 6.6 National olympiad Prove it

Lemma 9 According to the usual method, write a positive integer $a$ in the form of a decimal number, that is,
$a=a_{n} 10^{n}+a_{n-1} 10^{n-1}+\cdots+a_{0}, \quad 0 \leqslant a_{i}<10$
When 9 can divide $a_{n}+a_{n-1}+\cdots+a_{0}$ , then we have 9 can divide $a$ . And when 9 cannot divide $a_{n}+a_{n-1}+\cdots+a_{0}$ , then 9 cannot divide $a$ .

Solution

By (13) and Lemma 8, we have
$a \equiv a_{n}+a_{n-1}+\cdots+a_{0}(\bmod 9)$

When 9 can divide $a_{n}+a_{n-1}+\cdots+a_{0}$ , then by (17) we get that 9 can divide $a$ . And when 9 cannot divide $a_{n}+a_{n-1}+\cdots+a_{0}$ , then by (17) we get that 9 cannot divide $a$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.