Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎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 19:50, 6 January 2021

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.

Video Solution

https://youtu.be/Ji5BR4SFkeE

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_22&oldid=141633"