2013 AMC 12B Problems/Problem 15
- The following problem is from both the 2013 AMC 12B #15 and 2013 AMC 10B #20, so both problems redirect to this page.
Problem
The number is expressed in the form
where and are positive integers and is as small as possible. What is ?
Solution
The prime factorization of is . To have a factor of in the numerator, must equal . Now we notice that there can be no prime which is not a factor of 2013 such that because this prime will not be represented in the denominator, but will be represented in the numerator. The highest less than is , so there must be a factor of in the denominator. It follows that , so the answer is , which is . One possible way to express is
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.