Difference between revisions of "2012 AMC 10A Problems/Problem 17"
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Revision as of 11:04, 4 July 2013
Contents
Problem
Let and be relatively prime integers with and = . What is ?
Solution 1
Since and are both integers, so must and . For this fraction to simplify to , the denominator, or , must be a multiple of 3. Looking at the answer choices, it is only possible when .
Solution 2
Using difference of cubes in the numerator and cancelling out one in the numerator and denominator gives .
Set , and . Then . Cross multiplying gives , and simplifying gives . Since and are relatively prime, we let and , giving and . Since , the only solution is , which can be seen upon squaring and summing the various factor pairs of .
An alternate method of solving the system of equations involves solving the second equation for , plugging it into the first equation, and solving the resulting quartic equation with a substitution of . The four solutions correspond to
Thus, the desired quantity .
Note that if you double and double , you will get different (but not relatively prime) values for and that satisfy the original equation.
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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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