Prove that for $n\in\mathbb{N}$, $\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=15\iff\left[\frac{n}{5}\right]=13.$ How to show that the following relation? : for $n\in\mathbb{N}$, $$\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=15\iff\left[\frac{n}{5}\right]=13.$$ It's not obvious to me. Can anyone help me? Thank you!

It's simple.
Let us assume $\left[\frac{n}{5}\right]=13$ to be true.
Since $\left[\frac{n}{5}\right]=13$,
$\Rightarrow\frac{n}{5}\in[13,14)$
$\Rightarrow n\in[13*5,14*5)$
$\Rightarrow n\in[65,70)$
Let us consider,
$$\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=\left[\frac{n}{5}\right]+\left[\frac{n}{5^2}\right]+\left[\frac{n}{5^3}\right]+\left[\frac{n}{5^4}\right]+ …...$$
We know that $\left[\frac{n}{5}\right]=13$.
Previously we had found $n\in[65,70)$, so $\frac{n}{5^2}\in\left[\frac{65}{25},\frac{70}{25}\right)$, or $\frac{n}{5^2}\in\left[2.6,2.8\right)$
So, $\left[\frac{n}{5^2}\right]=2$
For other terms till infinity, the fraction inside the floor function goes below one and thus the value of floor will be zero. So we finally get the following result,
$$\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=13+2=15$$
Hence proved!
