Difference between revisions of "2012 AMC 10A Problems/Problem 17"
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So <math>a-b=\boxed{\textbf{(C)}\ 3}</math> | So <math>a-b=\boxed{\textbf{(C)}\ 3}</math> | ||
+ | == Solution 4 == | ||
+ | |||
+ | Slightly expanding, we have that | ||
+ | <math>\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}</math>. | ||
+ | Canceling the <math>(a-b)</math>, cross multiplying, and simplifying, we obtain that | ||
+ | <math>0=70a^2-149ab+70b^2</math>. | ||
+ | Dividing everything by <math>b^2</math>, we get that | ||
+ | <math>0=70(\frac{a}{b})^2-149(\frac{a}{b})+70</math>. | ||
+ | Applying the quadratic formula....and following the restriction that <math>a>b>0</math>, we get that | ||
+ | <math>\frac{a}{b}=\frac{10}{7}</math>. | ||
+ | Hence, <math>7a=10b</math>. | ||
+ | Since they are relatively prime, <math>a=10</math>, <math>b=7</math>. | ||
+ | <math>10-7=\boxed{\textbf{(C)}\ 3}</math>. | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:57, 4 February 2018
Contents
Problem
Let and
be relatively prime integers with
and
=
. What is
?
Solution 1 (The not so good solution)
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
, we get 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.