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

Revision as of 19:53, 2 November 2020

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.

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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Revision as of 17:47, 19 October 2020 (view source) Heinold200718 (talk \| contribs) (→‎Problem) ← Older edit		Revision as of 19:53, 2 November 2020 (view source) Mathfan335 (talk \| contribs) (→‎Problem) Newer edit →
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	== Problem ==		== Problem ==
−	~~How~~ many integers <math>n</math> ~~satisfy both of the inequalities~~ <math>~~4n + 3~~ < 25</math> ~~and~~ <math>~~-7n~~ + ~~5 < 24~~</math>?	+	For how many positive integers <math>n</math> less than or equal to <math>24</math> is <math>n!</math> evenly divisible by <math>1 + 2 + \cdots + n?</math>

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

Revision as of 19:53, 2 November 2020

Problem

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