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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.

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  • $\begingroup$ In other words, the number of zeros in the decimal representation of powers $5^n$ is unbounded. I see it as a difficult problem (very difficult?). $\endgroup$
    – Piquito
    Commented Feb 25 at 21:20
  • $\begingroup$ @Piquito , most likely it`s not, this is a task from my seminar. it can be noted that decimal places in the powers of 5 is cyclic (and length of the every cycle is power of 2). May be it can help. $\endgroup$
    – Email
    Commented Feb 25 at 22:03
  • $\begingroup$ See also oeis.org/A063585 and oeis.org/A329172 $\endgroup$
    – lhf
    Commented Feb 26 at 11:51

1 Answer 1

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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.

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