Difference between revisions of "2012 AMC 10A Problems/Problem 17"
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<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 (Quick Insight)== |
Since <math>a</math> and <math>b</math> are relatively prime, <math>a^3-b^3</math> and <math>(a-b)^3</math> are both integers as well. Then, for the given fraction to simplify to <math>\frac{73}{3}</math>, the denominator <math>(a-b)^3</math> must be a multiple of <math>3.</math> Thus, <math>a-b</math> is a multiple of <math>3</math>. Looking at the answer choices, the only multiple of <math>3</math> is <math>\boxed{\textbf{(C)}\ 3}</math>. | Since <math>a</math> and <math>b</math> are relatively prime, <math>a^3-b^3</math> and <math>(a-b)^3</math> are both integers as well. Then, for the given fraction to simplify to <math>\frac{73}{3}</math>, the denominator <math>(a-b)^3</math> must be a multiple of <math>3.</math> Thus, <math>a-b</math> is a multiple of <math>3</math>. Looking at the answer choices, the only multiple of <math>3</math> is <math>\boxed{\textbf{(C)}\ 3}</math>. |
Revision as of 19:16, 17 August 2023
Contents
Problem
Let and
be relatively prime positive integers with
and
. What is
?
Solution 1 (Quick Insight)
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 | |
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.