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Let $a$ be a three digit integer number with digits $x; y; z $(in that order). Prove that $a$ is divisible by 9 if and only if $x + y + z$ is divisible by 9.

Following a proof of this: Let $a; b; d; k$ be integers such that $a = dk + b$. Prove that $a$ is divisible by $d$ if and only if $b$ is divisible by $d$.

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    $\begingroup$ Well let $a=100x+10y+z$ with $0\leq x,y,z\leq 9$ and $x,y,z\in\mathbb{Z}$. The sum of its digits is $s=x+y+z$. Notice that $a-s=99x+9y=9(11x+y)\Leftrightarrow 9|(a-s)$. Can you finish? $\endgroup$ – JC12 Nov 10 '20 at 0:44
  • $\begingroup$ Does this answer your question? Show that the number 9 divides the number $ m$ if and only if the sum of the digits of the number $ m $ is divisible by 9. $\endgroup$ – John Omielan Nov 10 '20 at 0:49
  • $\begingroup$ Out of curiosity, why do you pick 100, 10, and 1 as the coefficients of x,y,z? $\endgroup$ – user825199 Nov 10 '20 at 0:53
  • $\begingroup$ The number $\overline {xyz}$ in base 10 evaluates to $100x + 10y + z$, the same way the number $3236$ in base 10 evaluates to $1000\times 3 + 100 \times 2 + 10 \times 3 + 6$. $\endgroup$ – player3236 Nov 10 '20 at 2:07
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Notice that the number $xyz$ equals $$100x+10y+z = 9(11x+y)+(x+y+z),$$ which is clearly divisible by $9$ if and only if $x+y+z$ is.

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