2005 AMC 10B Problems/Problem 22
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. ! Missing $ inserted.). 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. ! Missing $ inserted.) 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 |
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