Competition problem — Number theory

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Number theory Difficulty 4.7 AIME Prove it Estonia

Juku claims that if the sum of the squares of all digits of a natural number is divisible by $3$ then the number itself is divisible by $3$ . Is Juku's claim always true?

Solution

The sum of the squares of the digits of the number $112$ is $6$ which is divisible by $3$ , while the number $112$ is not divisible by $3$ .

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