2012 AMC 10A Problems/Problem 17
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 either be seen upon squaring and summing the various factor pairs of
.
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 | |
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All AMC 10 Problems and Solutions |