I am trying to solve this problem:
Prove that for every natural number $m$, there exists a natural number $n$ such that in the decimal representation of the number $5^n$ there are at least $m$ zeros.
Before that I proved that for $m$ = $2^a$ $(a \ge 3)$, we have $Z^*_{m} = \langle -1\rangle_2*\langle5\rangle_{2^{a-2}}$.
These tasks are on the topic of primitive roots, so how can I reformulate my problem in these terms? I think these tasks may be related to each other, but I don't see how exactly.
Give me some tips please.
I think there is a much simpler method. Please notice that:
$$10^n \mid 5^{2^{n-1}+n}-5^n.$$
Hence, there is $m \in \mathbb N$ such that:
$$m \times 10^n +5^n=5^{2^{n-1}+n}.$$
Now letting $n \to \infty$ implies that there are consecutive zeros in the middle of $5^{2^{n-1}+n}$ as many as required since $10^n$ grows faster than $5^n.$
We are done.