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

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.

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.

@@ Line 1: / Line 1: @@
 == 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>

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

Revision as of 17:47, 19 October 2020

Problem

Solution

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