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Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$

Attempt: I want to use the following theorem:

The largest exponent of $e$ of a prime $p$ such that $p^e$ is a divisor of $n!$ is given by $$e=[\frac{n}{p}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+\cdots$$.

The number of times the prime divisor $5$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $5$ contained in $(5^{25}-1)!$, which is $e_1=[\frac{(5^{25}-1)}{5}]+[\frac{(5^{25}-1)}{5^2}]+[\frac{(5^{25}-1)}{5^3}]+[\frac{(5^{25}-1)}{5^4}]+\cdots$

The number of times the prime divisor $2$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $2$ contained in $(5^{25}-1)!$, which is $e_2=[\frac{(5^{25}-1)}{2}]+[\frac{(5^{25}-1)}{2^2}]+[\frac{(5^{25}-1)}{2^3}+[\frac{(5^{25}-1)}{2^4}]+\cdots$

Therefore, the number of zeros at the right end equals the greatest exponent of $10$ contained in $(5^{25}-1)!$$=\min{(e_1, e_2)}$.

I am unable to simplify $e_1, e_2$. Please help me.

user1942348
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4 Answers4

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First in the définition of $e_1$ and $e_2$ there is no factoriel $!$ we have: $$ e_1=\left\lfloor\frac{5^{25}-1}{5}\right\rfloor+\left\lfloor\frac{5^{25}-1}{5^2}\right\rfloor+\left\lfloor\frac{5^{25}-1}{5^3}\right\rfloor+\left\lfloor\frac{5^{25}-1}{5^4}\right\rfloor+\cdots+\left\lfloor\frac{{5^{25}-1}}{5^{24}}\right\rfloor $$

And here you forgot to add the function : the greatest number less than $5^n-1$ each time because the numbers to sum must be all integers and finally we have to stop at $5^{24}$ because the rest are all null and we have: $$e_1=5^{24}-1+5^{23}-1+\dots+5^0-1=\frac{5^{25}-1}{4}-25=\frac{5^{25}-101}{4}$$

and it's clear that the greatest power of $5$ dividing $(5^{25}-1)!$ is the smallest one. each time we have a $5$ n=in the factoriel we will have more then $3$ powers of $2$, you can prove this by comparing the formula of $e_1$ and $e_2$ for instance here is an example $1\cdot 2\cdot 3\cdot 4\cdot 5=5\cdot 3\cdot 2^3$

Elaqqad
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As shown in this answer, the number of factors of a prime $p$ in $n!$ is $$ \frac{n-\sigma_p(n)}{p-1} $$ where $\sigma_p(n)$ is the sum of the digits in the base-$p$ expansion of $n$.

The base-$5$ expansion of $5^{25}-1$ consists of $25\,\,4$s, thus $$ \sigma_{5}(5^{25}-1)=25\cdot4=100 $$ Thus, the number of factors of $5$ in $(5^{25}-1)!$ is $$ \frac{5^{25}-1-100}{5-1}=\frac{5^{25}-101}{4} $$

The binary representation of $5^{25}-1$ has at most $3\cdot25=75$ bits since $5\lt2^3$. Thus, the number of factors of $2$ in $(5^{25}-1)!$ is at least $$ \frac{5^{25}-1-75}{2-1}=5^{25}-76\gt\frac{5^{25}-101}{4} $$

Thus, the number of factors of $10$ in $(5^{25}-1)!$ is $\dfrac{5^{25}-101}{4}$.

Community
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robjohn
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Since $e_1\le e_2$, all you need is $e_1$.

$$\begin{align}e_1&=\sum_{k=1}^{25}\left\lfloor\frac{5^{25}-1}{5^k}\right\rfloor\\&=\sum_{k=1}^{25}\left\lfloor 5^{25-k}-\frac{1}{5^k}\right\rfloor\\&=\sum_{k=1}^{25}5^{25-k}+\sum_{k=1}^{25}\left\lfloor-\frac{1}{5^k}\right\rfloor\\&=\frac{5^{25}-1}{4}+25\times(-1)\\&=\frac{5^{25}-101}{4}\end{align}$$

mathlove
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  • If I also want to find $e_2$ for some other need, then what will be the procedure? – user1942348 May 08 '15 at 09:34
  • $e_2=\sum_{k=1}^{\infty}\left\lfloor\frac{5^{25}-1}{2^k}\right\rfloor$, but this may be more difficult than the case of $5$, I think. As people point out, all we need is $e_1$. – mathlove May 08 '15 at 09:39
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Actually it would be: $$\small e_1=\sum_{k=1}^{\infty} \left\lfloor\frac{5^{25}-1}{5^k}\right\rfloor=\sum_{k=1}^{\infty} \left\lfloor5^{25-k}-\frac1{5^k}\right\rfloor=\sum_{k=1}^{25} (5^{25-k}-1)\tag{$\because \left\lfloor-\frac1{5^k}\right\rfloor=-1$}$$

RE60K
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