Lack of clarity over modular arithmetic notation

Question

While looking at a solution to a question, the following step is presented:

$5^n$ mod $10^k < 10^{k-6}$

$5^k$ ($5^{n-k}$ mod $2^k) < 10^{k-6}$

I don't quite understand how $5^k$ can be 'factored out' (which is my interpretation of what is happening here).

I can see that $10^k = 2^k 5^k$ but an explanation of why factoring is acceptable here would be appreciated

For anyone interested the question and solution is here: https://artofproblemsolving.com/wiki/index.php?title=2016_USAJMO_Problems/Problem_2

Answer 1

$a \equiv b*m \mod c*m$ means

$a = b*m + k(c*m) = m(b + kc)$ which means $m|a$ and

$\frac am = b + kc$ which means

$\frac am \equiv b \mod c$.

...

In this case they are abusing the notation but they mean

$(5^n \mod 10^k) = M = 5^n + K*10^k$ for some integer $M$ and some integer $K$ (REALLY bad abuse of notataion! .... but oh, well. That's the way the world is these days.....)

So $M = 5^k(5^{n-k} + K 2^k)$.

If we're allowed to say stupid things like $(5^n \mod 10^k)=M$ is a number $M$ then we are allowed to say a stupid thing like $5^{n-k} \mod 2^k = V$ is a number and cleary $5^k *V = M$ so...