Is there a way to figure out the $n$ such that $n!$ ends with exactly $k$ zeroes? Given $n$, I can find the number of zeroes at the end of the decimal representation of $n!$ by
$$
\sum_{i=1}^\infty\left\lfloor\frac{n}{5^i}\right\rfloor.
$$
Is there a way to reverse this? That is, given $k$, is there a way to find out how many $n$ exist such that $n!$ has exactly $k$ zeroes at the end of its decimal representation besides making educated guesses and checking them?
 A: There's a way to almost reverse this. I find it easier to think about this in terms of quinary (base $5$) representations. So your formula says that the number of zeroes in the decimal representation of $n!$ is the sum of all right-shifted and truncated versions of the quinary representation of $n$. That means that the $m$-th digit $a_m$ of the quinary representation of $n$ contributes
$$
\sum_{i=0}^{m-1}5^i=\frac{5^m-1}4
$$
zeroes, so we have
$$
4k=\sum_m a_m5^m-\sum_ma_m=n-\sum_ma_m\;.
$$
So $n$ is basically $4k$, and then you have to adjust a bit because the sum of the quinary digits gets subtracted.
A: Comparing the amount of information in OEIS A027868 and OEIS A007845, you may not be able to do much better than an educated guess.
Your first educated guess might be to look at $$\sum_{i=1}^\infty\frac{n}{5^i}=\frac{n}{4}.$$  But using $4k$ will be too low, because of rounding.  
How far too low? 
Note that if $k=776$ then $n=3120 = 4 \times 776 +16$ (or $3121,3122,3123,3124$) 
while if $k=781$ then $n=3125 = 4 \times 781+1$ (or the next four).    
If $4k+1$ is just below a multiple of a large power of $5$ then the amount you have to add on will be greater than if $4k+1$ is equal to or just above a multiple of a large power of $5$, but I suspect it may be quicker to do this by experiment than to seek a formula.  
A: This is wrong:
Well, a number is going to have as many 0's at the end as it has factors of 10 in it, and since $n!$ is going to have an abundance of 2's, you only need to look for 5's. So every time you hit a $n$ that is divisible by 5, $n!$ gains another 0. So there are 5 n such that $n!$ has k 0's. Specifically, $n = 5k,\, 5k+1,\, 5k+2,\, 5k+3,\, 5k+4$
At least I'm pretty sure I'm right, but I have no experience with number theory.
Why it is wrong: 25! has 2 5s, so does 50!, 75!; 125! has 3 etc. ie, Problems with higher powers of 5. But you will have either 5 or 0 n for any k.
