What is the maximum value of $n$ if $4^n$ divides $1000!$ without a remainder?

Question

If $1000!$ is divided by $4^n$ with a remainder 0, what is the highest possible value of $n$?

I placed 2, 3, 4, etc value in $n$ but didn't found any possible $4^n$. Moreover I have seen that only $4^1$ can divide 1000! without remainder. Is there any way to prove this or there are some possible $n$?

Answer 1

According to Legendre's Theorem (which is a very useful tool for such problems) $$\nu_p(n!)=\frac{n-s_p(n)}{p-1}$$ With $p=2$ and $n=1000$ $$\nu_2(1000!)=1000-s_2(1000)$$ Also $(1000)_{10}=(1111101000)_2$ so $s_2(1000)=1+1+1+1+1+0+1+0+0+0=6$, thus $$\nu_2(1000!)=1000-6=994$$ or $$2^{994}=4^{497} \mid 1000!$$

Answer 2

For a solution by hand: $1000!$ has

$500$ multiples of $2^{1}$

$250$ multiples of $2^{2}$

$125$ multiples of $2^{3}$

$62$ multiples of $2^{4}$

$31$ multiples of $2^{5}$

$15$ multiples of $2^{6}$

$7$ multiples of $2^{7}$

$3$ multiples of $2^{8}$

$1$ multiples of $2^{9}$

So $2^{1+3+7+15+...+500}=2^{994}=4^{497}|1000!$