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

m (Solution)
(Problem)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
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 + \ldots + n</math>?
+
How many integers <math>n</math> satisfy both of the inequalities <math>4n + 3 < 25</math> and <math>-7n + 5 < 24</math>?
 +
 
  
 
<math>\text{(A) 8    } \text{(B) 12    } \text{(C) 16    } \text{(D) 17    } \text{(E) 21    }</math>
 
<math>\text{(A) 8    } \text{(B) 12    } \text{(C) 16    } \text{(D) 17    } \text{(E) 21    }</math>

Revision as of 17:47, 19 October 2020

Problem

How many integers $n$ satisfy both of the inequalities $4n + 3 < 25$ and $-7n + 5 < 24$?


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

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.

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS