2019 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 25 Hard

For how many integers n between 1 and 50, inclusive, is \frac{(n^2-1)!}{(n!)^n} an integer? (Recall that 0!=1.) (2019 AMC 10A Problem, Question#25)

  • A.

    31

  • B.

    32

  • C.

    33

  • D.

    34

  • E.

    35

Answer:D

The main insight is that \frac{(n^2)!}{(n!)^{n+1}}

is always an integer. This is true because it is precisely the number of ways to split up n^2 objects into n unordered groups of size n. Thus,

\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}

is an integer if n^2\left| n!\right., or in other words, if n\left| (n-1)!\right.. This condition is false precisely

when n=4 or n is prime, by Wilson's Theorem. There are 15 primes between 1 and 50, inclusive, so there are 15 + 1 = 16 terms for which \frac{(n^2-1)}{n!}^n

is potentially not an integer. It can be easily verified that the above expression is not an integer for n=4 as there are more factors of 2 in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime n=p, as there are more factors of p in the denominator than the numerator. Thus all 16 values of n make the expression not an integer and the answer is 50-16=\boxed{(\text{D})34}.

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