Difference between revisions of "2005 AMC 10B Problems/Problem 22"
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− | \boxed{\textbf{(C) }16}</math> numbers less than or equal to 24 that satisfy the condition. | + | \boxed{\textbf{(C) }16}</math> numbers less than or equal to <math>24</math> that satisfy the condition. |
==Video Solution== | ==Video Solution== |
Revision as of 16:34, 16 December 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
odd primes less than or equal to
, so there
are numbers less than or equal to
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.