TAOCP 1.2.4 Exercise 15
By definition (1), x \bmod y = x-y\left\lfloor \frac{x}{y}\right\rfloor .
Section 1.2.4: Integer Functions and Elementary Number Theory
Exercise 15. [10] Prove that $z(x \bmod y) = (zx) \bmod (zy)$. [Law C is an immediate consequence of this distributive law.]
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By definition (1),
$$ x \bmod y = x-y\left\lfloor \frac{x}{y}\right\rfloor . $$
Hence
$$ z(x\bmod y) = z\left(x-y\left\lfloor \frac{x}{y}\right\rfloor \right) = zx-zy\left\lfloor \frac{x}{y}\right\rfloor . $$
Also,
$$ \left\lfloor \frac{zx}{zy}\right\rfloor = \left\lfloor \frac{x}{y}\right\rfloor , $$
provided that $zy\ne0$. Therefore
$$ (zx)\bmod(zy) =zx-zy\left\lfloor \frac{zx}{zy}\right\rfloor =zx-zy\left\lfloor \frac{x}{y}\right\rfloor =z(x\bmod y). $$
If $zy=0$, then either $z=0$ or $y=0$. When $z=0$,
$$ z(x\bmod y)=0=(0x)\bmod(0y). $$
When $y=0$,
$$ z(x\bmod0)=zx=(zx)\bmod0 $$
by definition (1). Thus the identity holds in all cases:
$$ \boxed{,z(x\bmod y)=(zx)\bmod(zy),}. $$