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

Question

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 $d$.

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? — JC12, Nov 10 '20 at 00:44
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. — John Omielan, Nov 10 '20 at 00:49
Out of curiosity, why do you pick 100, 10, and 1 as the coefficients of x,y,z? — , Nov 10 '20 at 00:53
The number $\overline {xyz}$ in base 10 evaluates to $100x + 10y + z$, the same way the number $3236$ in base 10 evaluates to $1000\times 3 + 100 \times 2 + 10 \times 3 + 6$. — player3236, Nov 10 '20 at 02:07

score 0 · Answer 1 · answered Nov 10 '20 at 00:53

0

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.

answered Nov 10 '20 at 00:53

Luke Collins

8,748

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

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