# What is the largest integer $k$ such that $4^{k} \vert 100!$?

What is the largest integer $k$ such that $4^k \vert 100!$. I understand the case where you have a composite number $n^k \vert 100!$ (where $n$ has two or more distinct primes), but I'm getting a little confused with the case where the prime appears more than once in the prime factorisation of $n$ (such as $4$). Thanks in advance.

Heres what I have so far...

$\lfloor{\frac{100}{2}}\rfloor = 50$, $\lfloor{\frac{100}{2^2}}\rfloor = 25$, $\lfloor{\frac{100}{2^3}}\rfloor = 12$, $\lfloor{\frac{100}{2^4}}\rfloor = 6$, $\lfloor{\frac{100}{2^5}}\rfloor = 3$, $\lfloor{\frac{100}{2^6}}\rfloor = 1$, $\lfloor{\frac{100}{2^7}}\rfloor = 0$

So the total number of factors of $2$ is equal to the sum of these numbers which is $97$.

• $4^k = (2^2)^k = 2^{2k}$. Now can you find the largest power of $2$ in $100!$ ? – Aritra Das May 22 '16 at 15:53
• Oh okay! So would I just combine these powers of $2$, i.e. compute $\lfloor{\frac{97}{2}}\rfloor$? – uk62116 May 22 '16 at 16:07
• use legendre's formula. – Jorge Fernández Hidalgo May 22 '16 at 16:07
• Alright thanks a lot! – uk62116 May 22 '16 at 16:10
• – N.S.JOHN May 22 '16 at 16:44