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 positive 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 (Quick Insight)== |
− | Since <math>a</math> and <math>b</math> are | + | 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>. |
== Solution 2 == | == Solution 2 == | ||
Line 34: | Line 34: | ||
Therefore <math>a=10</math> and <math>b=7</math> is a solution. | Therefore <math>a=10</math> and <math>b=7</math> is a solution. | ||
So <math>a-b=\boxed{\textbf{(C)}\ 3}</math> | So <math>a-b=\boxed{\textbf{(C)}\ 3}</math> | ||
+ | |||
+ | '''Note:''' | ||
+ | |||
+ | From <math>9ab=70(a-b)^2</math>, the Euclidean Algorithm gives <math>\gcd(a-b,a)=\gcd(a-b,b)=1</math>. Thus <math>(a-b)^2</math> is relatively prime to <math>ab</math>, and clearly <math>9</math> and <math>70</math> are coprime as well. The solution must therefore be <math>(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}</math> and <math>ab=70</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\left(\frac{a}{b}\right)^2-149\left(\frac{a}{b} \right)+70</math>. | ||
+ | |||
+ | Applying the quadratic formula....and following the restriction that <math>a>b>0</math>.... | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Note that the denominator, when simplified, gets <math>3.</math> 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 <math>\boxed{\textbf{(C)}\ 3}</math> ~mathboy282 | ||
+ | |||
+ | |||
+ | ==Solution 6== | ||
+ | Let us rewrite the expression as <math>\frac{(a-b)^2 + 3ab}{(a-b)^2}</math>. Now letting <math>x = a - b</math>, we simplify the expression to <math>\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}</math>. Cross multiplying and doing a bit of simplification, we obtain that <math>ab = \frac{70x^2}{9}</math>. Since <math>a</math> and <math>b</math> are both integers, we know that <math>\frac{70x^2}{9}</math> has to be an integer. Experimenting with values of <math>x</math>, we get that <math>x = 3</math> which means <math>ab = 70</math>. We could prime factor from here to figure out possible values of <math>a</math> and <math>b</math>, but it is quite obvious that <math>a = 10</math> and <math>b=7</math>, so our desired answer is <math>\boxed{\textbf{(C)}\ 3}</math> ~triggod | ||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | Since the two numbers are integers, and both <math>(a-b)^3</math> and <math>a^3-b^3</math> would yield integers, for the denominator to have a factor of 3, <math>(a-b)</math> must have a factor of 3. Only choice <math>\boxed{\textbf{(C)}}</math> has a factor of 3. ~hh99754539 | ||
+ | |||
+ | Note: isn't this solution (7) identical to solution #1?? | ||
+ | |||
+ | ==Solution 8== | ||
+ | We expand difference of cubes and cancel <math>a-b</math> from the numerator and denominator and see that <math>\dfrac{a^2-ab+b^2}{(a-b)^2}=\dfrac{73}3.</math> Obviously, we can not equate the numerator and denominator quite yet since that would imply that <math>a-b</math> is irrational (<math>\sqrt3</math>). We try the easiest thing to make <math>\dfrac{73}3</math>'s denominator a square: simply multiply by <math>\dfrac33,</math> giving <math>\dfrac{219}9.</math> Setting the denominators to be equal, we see that <math>(a-b)^2=9\implies a-b=3.</math> | ||
+ | |||
+ | ~ Technodoggo | ||
+ | |||
+ | == Video Solution by OmegaLearn== | ||
+ | https://youtu.be/ZWqHxc0i7ro?t=417 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/8SXVrlH71jk | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Latest revision as of 20:24, 14 June 2024
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
Note: isn't this solution (7) identical to solution #1??
Solution 8
We expand difference of cubes and cancel from the numerator and denominator and see that
Obviously, we can not equate the numerator and denominator quite yet since that would imply that
is irrational (
). We try the easiest thing to make
's denominator a square: simply multiply by
giving
Setting the denominators to be equal, we see that
~ Technodoggo
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.