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!
Best Answer
I think the easiest way to answer this question is to factorize $100!$. Actually, a partial factorization will be sufficient. Thus, we see that $$100! = 2^{97} \times 3^{48} \times 5^{24} \times \ldots \times 83 \times 89 \times 97.$$
Since $8 = 2^3$, we need to divide $97$ by $3$ and discard the remainder. That is, rewrite $2^{96}$ as $8^n$ and there's your answer.