Difference between revisions of "2005 AMC 10B Problems/Problem 22"

Revision as of 15:34, 16 December 2021

Problem

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

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

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 $24 - 8 = \boxed{\textbf{(C) }16}$ 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
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 are <math>24 - 8 =
-\boxed{\textbf{(C) }16}</math> numbers less than or equal to 24 that satisfy the condition.
+\boxed{\textbf{(C) }16}</math> numbers less than or equal to <math>24</math> that satisfy the condition.
 ==Video Solution==

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Difference between revisions of "2005 AMC 10B Problems/Problem 22"

Revision as of 15:34, 16 December 2021

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Problem

Solution

Video Solution

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