2005 AMC 10B Problems/Problem 22

Revision as of 23:16, 31 May 2021 by Mobius247 (talk | contribs) (Solution)

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}}$ (Error compiling LaTeX. ! Missing $ inserted.). 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}$ (Error compiling LaTeX. ! Missing $ inserted.) numbers less than or equal to 24 that satisfy the condition.

Video Solution

https://youtu.be/Ji5BR4SFkeE

~savannahsolver

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS