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!

Answers 2

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
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
Anurag A
August 14, 2019 05:17 AM

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