AMC 12 Mock Exam 3 Problem 17

By Legendre's formula, $v_2(m!)=\lfloor m/2\rfloor+\lfloor m/4\rfloor+\cdots=m-s(m)$, where $s(m)$ is the number of $1$'s in the binary representation of $m$. (Each $1$ in position $j$ contributes $2^{j-1}+2^{j-2}+\cdots+1=2^j-1$.) Since doubling does not change the binary digit sum, $$v_2\binom{2n}{n}=v_2((2n)!)-2v_2(n!)=\bigl(2n-s(n)\bigr)-2\bigl(n-s(n)\bigr)=s(n).$$ So $\binom{2n}n$ is not divisible by $4$ exactly when $s(n)\le1$, i.e., when $n$ is a power of $2$. For $1\le n\le100$ these are $1,2,4,8,16,32,64$: $\boxed{7}$ values.