2005 AMC 10B Problems/Problem 22

Problem

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

$\text{(A) 8 } \text{(B) 12 } \text{(C) 16 } \text{(D) 17 } \text{(E) 21 }$

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 24, so there are $24 - 8 = \boxed{\text{(C)}16}$ numbers less than or equal to 24 that satisfy the condition.

Video Solution

https://youtu.be/Ji5BR4SFkeE

~savannahsolver

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2005 AMC 10B Problems/Problem 22

Contents

Problem

Solution

Video Solution

See Also