Difference between revisions of "2005 AMC 10B Problems/Problem 22"
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== Solution == | == Solution == | ||
| − | Since <math>1 + 2 + \cdots + n = \frac{n(n+1)}{2}</math>, the condition is equivalent to having an integer value for <math>\frac{n!}{\frac{n(n+1)}{2}}</math>. This reduces, when <math>n\ge 1</math>, to having an integer value for <math>\frac{2(n-1)!}{n+1}</math>. This fraction is an integer unless <math>n+1</math> is an odd prime. There are 8 odd primes less than or equal to 24, so there are <math>24 - 8 = \boxed{\text{(C)}16}</math> numbers less than or equal to 24 that satisfy the condition. | + | Since <math>1 + 2 + \cdots + n = \frac{n(n+1)}{2}</math>, the condition is equivalent to having an integer value for <math>\frac{n!} |
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| + | {\frac{n(n+1)}{2}}</math>. This reduces, when <math>n\ge 1</math>, to having an integer value for <math>\frac{2(n-1)!}{n+1}</math>. This fraction is an | ||
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| + | integer unless <math>n+1</math> is an odd prime. There are 8 odd primes less than or equal to 24, so there are <math>24 - 8 = | ||
| + | |||
| + | \boxed{\text{(C)}16}</math> numbers less than or equal to 24 that satisfy the condition. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 22:16, 31 May 2021
Contents
Problem
For how many positive integers 
less than or equal to 
is 
evenly divisible by 
Solution
Since 
, the condition is equivalent to having an integer value for $\frac{n!}
{\frac{n(n+1)}{2}}$ (Error compiling LaTeX. Unknown error_msg). This reduces, when 
, to having an integer value for 
. This fraction is an
integer unless 
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. Unknown error_msg) numbers less than or equal to 24 that satisfy the condition.
Video Solution
~savannahsolver
See Also
| 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. 
