# 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!

• 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? – thewitness Aug 14 at 5:01

## 3 Answers

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 $$\left\lfloor\frac{n}{5}\right\rfloor=13.$$

• That some nice Latex'ing – Klangen Aug 14 at 13:33

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.

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!