Prove if the sum of the digits of a 3 digit number n is divisible by 9, then n is divisible by 9.

1 answer:

6 0

$n=100x+10y+z\\x+y+z=9k,k\in\mathbb{Z}\\\\n=99x+9y+x+y+z\\n=99x+9y+9k\\n=9(11x+y+k)$

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