Find $100$th number $k$ such that there is no $n$ for which $n$! ends in $k$ zeroes. 
$24! = 620,448,401,733,239,439,360,000$ ends in four zeroes, and $25! =
15,511,210,043,330,985,984,000,000$ ends in six zeroes. Thus, there is no integer
$n$ such that $n!$ ends in exactly five zeroes. Let $S$ be the set of all $k$ such that for
no integer $n$ does $n!$ end in exactly $k$ zeroes. If the numbers in $S$ are listed in
increasing order, $5$ will be the first number. Find the $100$th number in that list.

I used the approach of finding the number of $5$s in $(n+1)(n+2)...(2n)$ and $(n)(n-1)...1$ and their difference will be the number of zeroes in $\binom{2n}{n}$. But I'm still not sure how to find a pattern for the number of zeroes not possible for any $n$.
 A: The number of factors fives in (and hence number of zeroes at the end of) $n!$ is $$\lfloor \tfrac n5\rfloor+\lfloor \tfrac n{25}\rfloor+\lfloor \tfrac n{125}\rfloor+\cdots$$ (the number of twos is always big enough to not be problematic). This value is non-decreasing with $n$ and advances by $1$ or more exactly at multiples of $5$ because that's where $\lfloor \frac n5\rfloor$ jumps up by $1$. The "or more" happend each time we hit a multiple of $25$, when we jump by $2$ (or by $3$ if we are at a multiple fo $125$, or by $4$ if ...) This can help you count gaps.
A: Let $n!$ ends on $m(n)=\lfloor\frac{n}{5}\rfloor+\lfloor\frac{n}{25}\rfloor+\lfloor\frac{n}{125}\rfloor+\ldots$ zeroes. And let $k(n)=\lfloor\frac{n}{5}\rfloor$. Let call "zero jump" every situation when $a!$ has more zeroes than $(a-1)!$. Then for $a\leq n$ there are exactly $k(n)$ zero jumps. Then set of $m(a)$ for $a\leq n$ consists of $1+k(n)$ numbers, starting from 0. Then $m(n)-k(n)$ numbers are omitted. We need to find minimum $n$ when $m(n)-k(n)\geq 100$. $m(n)-k(n)=\lfloor\frac{n}{25}\rfloor+\lfloor\frac{n}{125}\rfloor+\ldots$. We can estimate this as $m(n)-k(n)\approx \frac{n}{25}+\frac{n}{125}+\ldots=\frac{n}{20}$. Consider case $n=20\cdot 100=2000$. Then $k(n)=400$, $m(n)=400+80+16+3=499$ which is 99 greater than $k(n)$. So we need to find next omitted number. $n=2005$, $n=2010$, $n=2015$, $n=2020$ all add 1 zero, then there are no omitted numbers here. $n=2025$ adds 2 zeroes. $m(2024)=m(2020)+4=503$, $m(2025)=m(2024)+2=505$. Then 100-th omitted number is $504$.
