What is the largest integer value of $n$ for which $8^n$ evenly divides $(100!)$? I know that this may be an unnecessary question, but I am a bit confused. The problem asks for the highest integer $n$ such that $8$ to the power of $n$ is divisible, evenly of course, by $100$. Now, I searched the site, and, in general, it seems that one can use floor function for a problem like this, but this seems to only work for prime numbers possibly. My process, which I realized was incorrect:
The floor function of $100/8 = 12$, and then doing it for the second power would lead to one, and, by adding those up, I acquired thirteen. Of course, after seeing the answer, $32$, I went back to see what was wrong and did the problem slower. I got $12$ numbers from the numbers in $100!$, and then got another $8$ from $2 \times 4$, but, that can be applied for all the multiples of $2$ and $4$ that aren't of $8$. So, essentially, I am wondering if there is a quicker method for calculating this number without specifically counting out the numbers. Thanks in advance!

It's easiest, I think, to do this with powers of $2:$
$$\left\lfloor{100\over2}\right\rfloor+
\left\lfloor{100\over4}\right\rfloor+
\left\lfloor{100\over8}\right\rfloor+
\left\lfloor{100\over16}\right\rfloor+
\left\lfloor{100\over32}\right\rfloor+
\left\lfloor{100\over64}\right\rfloor=97=32\cdot3+1
$$
so the greatest exponent of $8$ is $32$.
