Question

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

$8$

$12$

$16$

$17$

$21$

Solution:

Since $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},$ divisibility is equivalent to $\dfrac{n!}{n(n+1)/2} = \dfrac{2(n-1)!}{n+1}$ being an integer.

If $n + 1$ is not prime (and $n \ge 1$ ), its factors appear among $1, 2, \ldots, n-1$ or in the factor $2,$ so the fraction is an integer. If $n + 1$ is an odd prime, it divides neither $(n-1)!$ nor $2,$ so the fraction is not an integer.

The odd primes at most $25$ are $3, 5, 7, 11, 13, 17, 19, 23,$ giving $8$ failing values of $n.$ Hence $24 - 8 = 16$ values work.

Thus, C is the correct answer.

2005 AMC 10B Problem 22

Solution:

Problem 22 in Other Years