Express $n=10k+d$. i.e $d$ is the last digit of $n$.
I am wondering how I can prove this criterion
$17|n $ if and only if $17|(3k+2d)$
I can prove the sufficient condition: $17|n$ iff $-10k\equiv d$ $(mod$ $17)$ , then $2d+3k\equiv2(-10k)+3k\equiv-17\equiv0$ $(mod$ $17)$.
But I got stuck proving the another side.
Observe that by eliminating one of the variables
$$10(3k+2d)-3(10k+d)=17d$$
$$2(10k+d)-(3k+2d)=17k$$
As the right hand side is divisible by $17,$
$$17\mid(3k+2d)\iff17|(10k+d)$$
$$n=10k+d$$
$$10k+d \equiv 0 \pmod{17}$$
Multiplying by $2$,
$$20k+2d\equiv 3k+2d \equiv 0\pmod{17}$$
Conversely, notice that $2$ and $17$ are coprime
If $3k+2d \equiv 0 \pmod{17}$, multiply both sides by $2^{-1}\pmod{17}$
$3(2^{-1})\equiv 10\pmod{17}$, we have $10k+d \equiv n \equiv 0 \pmod{17}$.
$(*)$ Because $\gcd(2, 17) = 1$.
$(**)$ Because $20\equiv 3 \pmod{17}$.
$$\begin{align}{\rm Note}\,\ 17\mid\:\! y-10x\iff&\, 17\mid 2y-3x,\ \ {\rm by}\\[.3em] \bmod\, 17\!:\ y\equiv 10x\iff &\,2y\equiv \color{#c00}3x,\ \ \,\rm by\\[.3em] y\equiv 10x\iff &\,2y\equiv \color{#c00}{20}x\ \ \,{\rm by}\ \ \color{#c00}{20\equiv 3} \end{align}\qquad$$
$(\Rightarrow)$ follows by scaling $\,y\equiv 10x\,$ by $2$. This scaling is is invertible: scale $\,2y\equiv \color{#c00}3x\,$ by $\,\frac{1}2 \equiv \frac{18}2\equiv 9$ to get $(\Leftarrow).\,$ Generally scaling equations by units (invertibles) yields equivalant equations (for our case of $\,\Bbb Z_m = $ integers $\!\bmod m,\,$ recall by Bezout $\,a\,$ is invertible $\!\bmod m\iff \gcd(a,m)=1)$.
Said fractionally $\bmod 17\!:\ 10\equiv 3/2\,$ by $\,10\cdot 2\equiv 3.\,$ Generally such divisibility tests employ such modular fractions for $10$ (or $1/10$) that are "small", so to make the divisibility test convenient for hand calculation.
Beware $ $ Modular fractions $\,a/b := ab^{-1}$ are well-defined only for fractions whose denominator $\,b\,$ is coprime to the modulus. See here for further discussion, and see here and here for many ways to calculate modular fractions and inverses.
