Prove that if $N =\overline {xyz}\in\boldsymbol N$ (natural number) and if $x+y+z=9$ then $9$ divides $N$.

Question

Prove that if $N =\overline {xyz}\in\boldsymbol N$ (natural number) and if $x+y+z=9$ then $9$ divides $N$.

My try:

$100x+10y+z=(99+1)x+(9+1)x+(10-9)z=99x+x+9y+y+10z-9z=9(11x+y-z)+x+y+10z$

Then I can't, help me please.

Answer 1

$$\begin{align} \color{blue}{x+y+z=9}&\iff \color{red}{99x}+\color{blue}x+\color{red}{9y}+\color{blue}y+\color{blue}z=\color{blue}9+\color{red}{99x+9y} \\\,\\ &\therefore\,\, 100x+10y+z=9(11x+y+1) \end{align}$$ Since $x,y\in\Bbb N$, then $11x+y+1=k\in\Bbb N$.

Thus: $100x+10y+z=9k$ which means that $N$ is divisible by $9$.

Answer 2

$\rm mod\ 9\!:\ 10\equiv 1\Rightarrow 10^k\!\equiv 1\,\Rightarrow\, c_n10^n\!+\cdots+c_1 10+c_0 \equiv c_n+\cdots+c_1+c_0$

Or, written in polynomial language, $\,\ 10\equiv 1\,\Rightarrow\, p(10) \equiv p(1) = $ sum of digits.