2005 AMC 10B Problems/Problem 22

Problem

For how many positive integers n less than or equal to $24$ is $n!$ evenly divisible by $1 + 2 + \ldots + n$ ?

Solution

Since $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ , the condition is equivalent to having an integer value for $\frac{n!}{\frac{n(n+1)}{2}}$ . This reduces, when $n\ge 1$ , to having an integer value for $\frac{2(n-1)!}{n+1}$ . This fraction is an integer unless $n+1$ is an odd prime. There are 8 odd primes less than or equal to 25, so there are $24 - 8 = \boxed{16}$ numbers less than or equal to 24 that satisfy the condition.

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 10 Problems and Solutions

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2005 AMC 10B Problems/Problem 22

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