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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!

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    $\begingroup$ 15=sigma([n/5^k])<sigma(n/5^k).(strict inequality here) Sum the infinite GP to get n>60 or n/5>12. Now you are left with cases [n/5]=12,13,14. Can you handle this? $\endgroup$ – thewitness Aug 14 at 5:01
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Observe that $\sum_{k=1}^{\infty}\left\lfloor\frac{n}{5^k}\right\rfloor=\nu_5(n!)$, where $\nu_5(n!)$ is the exponent of the largest power of $5$ that divides $n!$ (if you are not familiar with this, then see Legendre's formula).

Since this sum is $15$, thus the highest power of $5$ that divides $n!$ is $15$.

Consider $n!=(1\cdot 2 \cdot 3 \cdot 4 \cdot \color{red}{5})(6\cdot 7 \cdot 8 \cdot 9 \cdot \color{red}{10})\dotsb ((n-4) \cdot (n-3) \cdot (n-2) \cdot (n-1) \cdot \color{red}{n}).$

Each group of $((a+1)\cdot (a+2) \cdot (a+3) \cdot (a+4) \cdot (a+5))$ contributes a single power of $5$, until we reach $25$, where we get a contribution of $2$ towards the exponent of $5$. Likewise until we reach $50$ each such group of five numbers contribute only a single power of $5$ and so on. Thus the number $n$ must be such that $$n!=\underbrace{(1\cdot 2 \cdot 3 \cdot 4 \cdot \color{red}{5})}_{5^1}\underbrace{(6\cdot 7 \cdot 8 \cdot 9 \cdot \color{red}{10})}_{5^1}\dotsb \underbrace{(21\cdot 22 \cdot 23 \cdot 24 \cdot \color{red}{25})}_{\color{green}{\boxed{5^2}}}\dotsb \underbrace{(61\cdot 62 \cdot 63 \cdot 64 \cdot \color{red}{65})}_{5^1} \dotsb ()$$ When $n=65$, that is the first time we will have $\nu_5(n!)=15$. Thus $65 \leq n < 70$ which implies $\left\lfloor\frac{n}{5}\right\rfloor=13.$

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    $\begingroup$ That some nice Latex'ing $\endgroup$ – Klangen Aug 14 at 13:33
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There are three parts to the claim:

  1. If $\frac{n}{5}<13$ then $\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]<15$
  2. If $13\leq\frac{n}{5}<14$ then $\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=15$
  3. If $\frac{n}{5}\geq14$ then $\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]>15$

So choose whichever part strikes your fancy and tackle it in isolation. Repeat until all three are done.

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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!

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