Difference between revisions of "2005 AMC 10B Problems/Problem 22"
Mathfan335 (talk | contribs) (→Problem) |
|||
Line 7: | Line 7: | ||
== 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!}{\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. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Ji5BR4SFkeE | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2005|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:50, 6 January 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
. 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
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.