2005 AMC 10B Problems/Problem 22
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
Video Solution
https://www.youtube.com/watch?v=XQQpjNuOL5E ~David
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
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Followed by Problem 23 | |
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