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For what $n$ the following is true: $$\left[\dfrac{n}{5}\right]+\left[\dfrac{n}{5^2}\right]+\left[\dfrac{n}{5^3}\right]+\cdots=100$$

My sketch: We know that $[x]\leq x<[x]+1$ for any real $x$. Also its obvious that above sum is finite. We can estimate this sum: $$100\leq \dfrac{n}{5}+\dfrac{n}{5^2}+\cdots=\sum \limits_{k=1}^{\infty}\dfrac{n}{5^k}=n/4$$ Hence $n>400$.

I check this out on the computer and yielded that it's true for $n\in \{405, 406, 407, 408,409\}$. Why my estimate is wrong?

S.C.B.
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RFZ
  • 16,814

3 Answers3

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Your estimate is not wrong. You estimated $n>400$. And it is true for $n=405, 406, 407, 408, 409$.

They all satisfy $n>400$.

S.C.B.
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1

Your estimate is not accurate because:

You should change this:

$$\color\red{100\leq\sum\limits_{k=1}^{\infty}\dfrac{n}{5^k}=n/4}$$

To this:

$$\color\green{100=\sum\limits_{k=1}^{\log_5n}\left\lfloor\dfrac{n}{5^k}\right\rfloor<\sum\limits_{k=1}^{\log_5n}\dfrac{n}{5^k}<\sum\limits_{k=1}^{\infty}\dfrac{n}{5^k}=n/4}$$

Keep in mind that we are dealing with positive integers here.

Therefore $k>\log_5n\implies5^k>n\implies\dfrac{n}{5^k}<1\implies\left\lfloor\dfrac{n}{5^k}\right\rfloor=0$.

barak manos
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  • I already meant this but I forgot mention it. Anyway your remark does not lead to a correct estimate. Am I right? – RFZ Feb 16 '17 at 08:06
  • @RFZ: You are trying to find the minimum value of $n$ such that the multiplicity of $5$ in $n$ is $100$, right? The formula for that is $\sum\limits_{k=1}^{\log_5n}\left\lfloor\frac{n}{5^k}\right\rfloor$, not $\sum\limits_{k=1}^{\infty}\frac{n}{5^k}$. – barak manos Feb 16 '17 at 08:06
  • @RFZ: No, it does not. It merely tries to explain why your estimate of $n=400$ is not accurate. – barak manos Feb 16 '17 at 08:07
  • Thanks a lot! It became much better – RFZ Feb 16 '17 at 08:13
  • @RFZ: You're welcome. I've expanded the explanation a little bit. – barak manos Feb 16 '17 at 08:16
  • That's great! Thank you :) – RFZ Feb 16 '17 at 08:18
1

From the initial estimate $n=400$, known to be a lower bound, write

$$\left[\dfrac{400+m}{5}\right]+\left[\dfrac{400+m}{5^2}\right]+\left[\dfrac{400+m}{5^3}\right]+\cdots=80+\left[\dfrac{m}{5}\right]+16+\left[\dfrac{m}{5^2}\right]+3+\left[\dfrac{25+m}{5^3}\right]+\left[\dfrac{400+m}{5^4}\right]++\left[\dfrac{400+m}{5^5}\right]+\cdots=100.$$

So you need

$$\left[\dfrac{m}{5}\right]+\left[\dfrac{m}{5^2}\right]+\left[\dfrac{25+m}{5^3}\right]+\left[\dfrac{400+m}{5^4}\right]+\left[\dfrac{400+m}{5^5}\right]+\cdots=1,$$

which is achieved for $\left[\dfrac m5\right]=1$, by inspection (for all $m$ such that $\left[\dfrac m5\right]\le1$, all terms but the first vanish).

  • You third term seems suspicious since $\left[\dfrac{400+m}{5^3}\right]=\left[\dfrac{375+25+m}{5^3}\right]=3+\left[\dfrac{25+m}{5^3}\right]$. Right? – RFZ Feb 16 '17 at 08:17
  • Quite right, fixing. This doesn't change the conclusion. By the way, I am giving a complete solution. –  Feb 16 '17 at 09:06