# identity for greatest integer function

by Primavera   Last Updated August 14, 2019 05:20 AM

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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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.

Chris Culter
August 14, 2019 05:01 AM

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

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

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 $$n!=\underbrace{(1\cdot 2 \cdot 3 \cdot 4 \cdot \color{red}{5})}_{5^1}\underbrace{(6\cdot 7 \cdot 8 \cdot 9 \cdot \color{red}{10})}_{5^1}\dotsb \underbrace{(21\cdot 22 \cdot 23 \cdot 24 \cdot \color{red}{25})}_{\color{green}{5^2}}\dotsb \underbrace{(61\cdot 62 \cdot 63 \cdot 64 \cdot \color{red}{65})}_{5^1}.$$ Thus $$n=65$$ which implies $$\left\lfloor\frac{n}{5}\right\rfloor=13.$$

Anurag A
August 14, 2019 05:17 AM

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