Question 41
For a positive integer $n$, let $f(n)$ be the largest $k$ such that $2^k$ divides $n$ and $g(n)$ be the sum of the digits of the binary representation of $n$. Prove that for any positive integers $n$,
(i) $f(n!) = n - g(n)$;
(ii) $4$ divides $\binom{2n}{n} = \frac{(2n)!}{n!n!}$ if and only if $n$ is not a power of $2$.
(i) $f(n!) = n - g(n)$;
(ii) $4$ divides $\binom{2n}{n} = \frac{(2n)!}{n!n!}$ if and only if $n$ is not a power of $2$.