2012 AMC 10A Problems/Problem 17
Contents
Problem
Let and be relatively prime positive integers with and . What is ?
Solution 1
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 .
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 .
Thus, .
Remarks:
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.
Solution 3
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
Note:
From , the Euclidean Algorithm gives . Thus is relatively prime to , and clearly and are coprime as well. The solution must therefore be and .
Solution 4
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 ....
.
Hence, .
Since they are relatively prime, , .
.
Solution 5
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
Solution 6
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 values of and , but it is quite obvious that and , so our desired answer is ~triggod
Solution 7
Since the two numbers are integers, and both and would yield integers, for the denominator to have a factor of 3, must have a factor of 3. Only choice has a factor of 3. ~hh99754539
Video Solution by OmegaLearn
https://youtu.be/ZWqHxc0i7ro?t=417
~ pi_is_3.14
Video Solution
~savannahsolver
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AMC 10 Problems and Solutions |
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