Difference between revisions of "2012 AMC 10A Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | Let <math>a</math> and <math>b</math> be relatively prime integers with <math>a>b>0</math> and <math>\ | + | Let <math>a</math> and <math>b</math> be relatively prime integers with <math>a>b>0</math> and <math>\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.</math> What is <math>a-b?</math> |
− | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5</math> | + | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5</math> |
==Solution 1== | ==Solution 1== |
Revision as of 10:54, 8 September 2020
Problem
Let and
be relatively prime 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
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, ,
.
.
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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