For how many integers n between 1 and 50, inclusive, is ((n^2 - 1)!)/((n!)^n) an integer? (Recall that 0! = 1. )

2019 AMC 12A Problem 24 — Solution

2019 AMC 12A Problem 24

Below is the professionally curated solution for Problem 24 of the 2019 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12A solutions, or check the answer key.

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Concepts:Legendre’s Formula prime digits

Difficulty rating: 2420

24.

For how many integers $n$ between $1$ and $50,$ inclusive, is $\dfrac{(n^2 - 1)!}{(n!)^n}$ an integer? (Recall that $0! = 1.$ )

$31$

$32$

$33$

$34$

$35$

Solution:

For a prime $p,$ the condition to be an integer reduces (via Legendre's formula) to $n \cdot s_p(n) \ge s_p(n^2 - 1) + 1,$ where $s_p$ is the base- $p$ digit sum. This can fail only when $s_p(n) = 1,$ i.e. $n = p^a$ is a prime power.

For $n = p^a,$ the requirement becomes $p^a - 1 \ge 2a(p - 1).$ Checking prime powers up to $50,$ this fails exactly for every prime $n$ and for $n = 4.$

There are $15$ primes at most $50,$ plus $n = 4,$ giving $16$ failures. Hence $50 - 16 = 34$ values of $n$ work.

Thus, the correct answer is D.

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