Difference between revisions of "2012 AMC 10A Problems/Problem 17"
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Note that if you double <math>x</math> and double <math>y</math>, you will get different (but not relatively prime) values for <math>a</math> and <math>b</math> that satisfy the original equation. | Note that if you double <math>x</math> and double <math>y</math>, you will get different (but not relatively prime) values for <math>a</math> and <math>b</math> that satisfy the original equation. | ||
+ | |||
+ | Or we can solve for <math>a-b</math> directly instead of solving for <math>a</math> and <math>b</math>. <math>a^2-2ab+b^2=149-140=9</math>, <math>a-b=3</math> | ||
== Solution 3 == | == Solution 3 == |
Revision as of 12:42, 1 December 2013
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.
Or we can solve for directly instead of solving for
and
.
,
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.