# number of 0 digits in rightward of $n!$ exactly equal is $[\frac n4]$

if $n$ is very big how to prove that number of 0 digits in rightward of $n!$ exactly equal is $[\frac n4]$ it's mean : $\lim_{n\to \infty} \frac{E_5(n!)}{[\frac n4]} =1$

that $E_5(n!)= \sum_{k=1}^\infty [\frac{n}{5^k}]$

it's proved that that number of 0-digits in rightward of $n!$ is $E_5(n!)$

I tried to prove it by sandwich theorem.

$E_5(n!) \le \frac n5+\frac {n}{5^2}+...=n\frac{\frac 15}{1-\frac 15}= \frac n4$ rightward of sandwich theorem is provided. leftward? how?

$$(??\le \frac{E_5(n!)}{[\frac n4]} \le 1 )$$

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What does "in rightward of $n!$" mean? (haven't seen that term.) – coffeemath Mar 16 '13 at 20:51
it means the largest power of 10 that divides n!. The number of trailing zeroes. – Carry on Smiling Mar 16 '13 at 21:21

I think you might want to show that: $\frac{[n/4]-C(n)}{[n/4]}\le \frac{E_5(n!)}{[n/4]} \le 1$, where lim $\frac{C(n)}{[n/4]} =0$ and by taking the limit, you can use the Sandwich method.

In order to prove the above, you need to show $E_5(n!) + C(n) \ge [\frac{n}{4}]$

It is equivalent to saying that $C(n) + [n/5]+[n/5^2]+....\ge [n/5+n/5^2+...]$ (*) So you need to find C(n) so when you add it to $E_5(n!)$, you have the above.

There is a fact, which is not hard to prove: $1+[x_1]+[x_2]\ge[x_1+x_2]$. From this, you can generalize it to more variables: $k-1+[x_1]+[x_2]+...+[x_k]\ge [x_1+x_2+...+x_k]$

In the (*) above, look at the left side, there are only a finite number K of terms of the forms $[n/5^k]$ which are different from 0. This K is ~ $log_5$ n. Using the inequality, you have:

K-1 + $[n/5]+[n/5^2]+...+[n/5^K]\ge[n/5+n/5^2+..+n/5^K]$

The tail: $n/5^{K+1}+n/5^{K+2}+...= n/5^{K+1} * (\frac{1}{1-1/5}) < 1 * 5/4 = 5/4$

From here, you have: 3+ (K-1) + $[n/5]+[n/5^2]+...+[n/5^K]\ge[n/5+n/5^2+..+n/5^K+....]$ So take C(n) = K+2, and you are done.

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As discussed here, the number of factors of $5$ in $n!$ is $$\frac{n-\sigma_5(n)}{4}\tag{1}$$ where $\sigma_5(n)$ is the sum of the base $5$ digits of $n$. Since the number of factors of $2$ that divide $n!$ is about 4 times the number of factors of $5$, there will be an ample number to match the $5$s. For example, in $3124!$ there are $776$ zeros to the right, whereas $\frac{3124}{4}=781$.

Now $\sigma_5(n)\le4(\lfloor\log_5(n)\rfloor+1)$ (the maximum digit times the number of digits); therefore, $$\frac n4-\lfloor\log_5(n)\rfloor-1\le\frac{n-\sigma_5(n)}{4}\le\frac n4\tag{2}$$ Since $$\lim_{n\to\infty}\frac{\log_5(n)}{n}=0\tag{3}$$ the Squeeze Theorem and $(2)$ and $(3)$ give your result.

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You guys are doing too much work...

In order for $n!$ to have a zero at the end, it must include a number of 2's and 5's. There are more 2's than there are 5's for any factorial, so counting the number of 5's in $n!$ will tell you the number of 0's at the end of $n!$.

So the numbers of 5's will be represented by $⌊\frac{n}{5}⌋$. However, this doesn't include the extra 5's from other powers of 5. like 25. The number of extra 5's would be $⌊\frac{n}{25}⌋$.

Continuing this pattern infinitely is simply a geometric series $$\frac{\frac{1}{5}n}{1-\frac{1}{5}} = \frac{1}{4}n$$

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This is not exact; as I computed above, $3124!$ has $776$ zeros on the right, $5$ fewer than $\frac{3124}{4}=781$. This error needs to be controlled to get the limit. – robjohn Mar 17 '13 at 1:47

Suppose that the base five representation of $n$ is $$n = \sum_{k=0}^{\lfloor \log_5 n\rfloor} d_k 5^k.$$ Then it is not difficult to see that $$E_5(n!) = -n + \sum_{k\ge 0} \Big\lfloor \frac{n}{5^k} \Big\rfloor = -n + \sum_{j=0}^{\lfloor \log_5 n\rfloor} \sum_{k=j}^{\lfloor \log_5 n\rfloor} d_k 5^{k-j}.$$ Let $$T(n) = \sum_{j=0}^{\lfloor \log_5 n\rfloor} \sum_{k=j}^{\lfloor \log_5 n\rfloor} d_k 5^{k-j}.$$ Now clearly the minimum for $T(n)$ is attained when $n$ consists of a single one digit followed by zeros, giving $$T(n) \ge \sum_{j=0}^{\lfloor \log_5 n\rfloor} 5^{\lfloor \log_5 n\rfloor-j} = 5^{\lfloor \log_5 n\rfloor} \sum_{j=0}^{\lfloor \log_5 n\rfloor} 5^{-j} = \frac{1}{4} 5^{\lfloor \log_5 n\rfloor+1} - \frac{1}{4} = \frac{5}{4} n \left(5^{\lfloor \log_5 n\rfloor-\log_5 n} - \frac{1}{5n}\right) = \frac{5}{4} n \left(1 - \frac{1}{5n}\right).$$ The last equality occurs because for $n$ as described, the floor of the logarithm stays exact.

The maximum is attained when all digits are equal to four, giving $$T(n) \le \sum_{j=0}^{\lfloor \log_5 n\rfloor} 4 \sum_{k=j}^{\lfloor \log_5 n\rfloor} 5^{k-j} = \frac{1}{4} 5^{\lfloor \log_5 n\rfloor+2} -\lfloor \log_5 n\rfloor - \frac{9}{4} = \frac{5}{4} n \left(5^{\lfloor \log_5 n\rfloor+1-\log_5 n} -\frac{4}{5}\frac{\lfloor \log_5 n\rfloor}{n} - \frac{9}{5n}\right).$$

Now the asymptotics require some work here. With all digits equal to four, we have $$n= 5^{\lfloor \log_5 n\rfloor+1} -1.$$ It follows that $$5^{\lfloor \log_5 n\rfloor+1-\log_5 n} = 5^{\lfloor \log_5 n\rfloor+1-\log_5 \left(5^{\lfloor \log_5 n\rfloor+1} -1\right) } = \frac{5^{\lfloor \log_5 n\rfloor+1}}{5^{\lfloor \log_5 n\rfloor+1}-1} \\ = \frac{1}{1-1/5^{\lfloor \log_5 n\rfloor+1}} = 1 + \left(\frac{1}{5^{\lfloor \log_5 n\rfloor+1}}\right) + \left(\frac{1}{5^{\lfloor \log_5 n\rfloor+1}}\right)^2 + \left(\frac{1}{5^{\lfloor \log_5 n\rfloor+1}}\right)^3 + \cdots.$$ This yields the following asymptotic expansion for the upper bound: $$T(n) \le \frac{5}{4} n \left(1 + \left(\frac{1}{5^{\lfloor \log_5 n\rfloor+1}}\right) + \left(\frac{1}{5^{\lfloor \log_5 n\rfloor+1}}\right)^2+\cdots -\frac{4}{5}\frac{\lfloor \log_5 n\rfloor}{n} - \frac{9}{5n}\right).$$ Finally, taking the upper bound and the lower bound together we have shown that $$T(n) \sim \frac{5}{4} n$$ and hence $$E_5(n!) \sim \frac{1}{4} n.$$

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