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

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 0:44
• Out of curiosity, why do you pick 100, 10, and 1 as the coefficients of x,y,z? – user825199 Nov 10 '20 at 0:53
• 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 2:07

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.