2012 AMC 10A Problems/Problem 17
Let and be relatively prime positive integers with and What is
Since and are relatively prime, and are both integers as well. Then, for the given fraction to simplify to , the denominator must be a multiple of Thus, is a multiple of . Looking at the answer choices, the only multiple of is .
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 , by plugging it into the first equation, and solving the resulting quartic equation with a substitution of . The four solutions correspond to
Also, we can solve for directly instead of solving for and :
Note that if you double and double , you will get different (but not relatively prime) values for and that satisfy the original equation.
The first step is the same as above which gives .
Then we can subtract and then add to get , which gives . . Cross multiply . Since , take the square root. . Since and are integers and relatively prime, is an integer. is a multiple of , so is a multiple of . Therefore and is a solution. So
From , the Euclidean Algorithm gives . Thus is relatively prime to , and clearly and are coprime as well. The solution must therefore be and .
Slightly expanding, we have that .
Canceling the , cross multiplying, and simplifying, we obtain that
. Dividing everything by , we get that
Applying the quadratic formula....and following the restriction that ....
Since they are relatively prime, , .
Note that the denominator, when simplified, gets We now have to test the answer choices. If one has a good eye or by simply testing the answer choices the answer will be clearly ~mathboy282
Let us rewrite the expression as . Now letting , we simplify the expression to . Cross multiplying and doing a bit of simplification, we obtain that . Since and are both integers, we know that has to be an integer. Experimenting with values of , we get that which means . We could prime factor from here to figure out possible vlaues of and , but it is quite obvious that and , so our desired answer is
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