Competition problem — Number theory

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Number theory Difficulty 5.5 AIME, harder Find the answer

23. Let $p$ be a prime number, and $n$ be a positive integer.
(i) Find the formula for $e$ in $p^{e} \|(2 n)!$ , where
$(2 n)!!=(2 n)(2 n-2) \cdots 2$
(ii) Find the formula for $f$ in $p^{f} \|(2 n+1)!!$ , where
$(2 n-1)!!=(2 n-1)(2 n-3) \cdots 1$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

23. (i) When $p=2$ , $e=n+\sum_{j}\left[n / 2^{j}\right] ;$ when $p>2$ , $e=\sum_{j}\left[n / p^{j}\right]$ .
(ii) When $p=2$ , $f=0 ;$ when $p>2$ , $f=\sum_{j}\left(\left[2 n / p^{j}\right]-\left[n / p^{j}\right]\right)$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.