Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2012 AMC 10A Problems/Problem 17: Difference between revisions

← Older editNewer edit →
Nathan wailes (talk | contribs)
2,960 edits
No edit summary
No edit summary
Line 5: Line 5:
<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 ==


Since <math>a</math> and <math>b</math> are both integers, so must <math>a^3-b^3</math> and <math>(a-b)^3</math>. For this fraction to simplify to <math>\frac{73}{3}</math>, the denominator, or <math>a-b</math>, must be a multiple of 3. Looking at the answer choices, it is only possible when <math>a-b=\boxed{\textbf{(C)}\ 3}</math>.
Since <math>a</math> and <math>b</math> are both integers, so must <math>a^3-b^3</math> and <math>(a-b)^3</math>. For this fraction to simplify to <math>\frac{73}{3}</math>, the denominator, or <math>a-b</math>, must be a multiple of 3. Looking at the answer choices, it is only possible when <math>a-b=\boxed{\textbf{(C)}\ 3}</math>.


== Solution 2==
== Solution 2 ==


Using difference of cubes in the numerator and cancelling out one <math>(a-b)</math> in the numerator and denominator gives <math>\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}</math>.
Using difference of cubes in the numerator and cancelling out one <math>(a-b)</math> in the numerator and denominator gives <math>\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}</math>.
Line 22: Line 22:


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.
== Solution 3 ==
The first step is the same as above which gives <math>\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}</math>.
Then we can subtract <math>3ab</math> and then add <math>3ab</math> to get <math>\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}</math>, which gives <math>1+\frac{3ab}{(a-b)^2}=\frac{73}{3}</math>. <math>\frac{3ab}{(a-b)^2}=\frac{70}{3}</math>.
Cross multiply <math>9ab=70(a-b)^2</math>. Since <math>a>b</math>, take the square root. <math>a-b=3\sqrt{\frac{ab}{70}}</math>.
Since <math>a</math> and <math>b</math> are integers and relatively prime, <math>\sqrt{\frac{ab}{70}}</math> is an integer. <math>ab</math> is a multiple of <math>70</math>, so <math>a-b</math> is a multiple of <math>3</math>.
Therefore <math>a=10</math> and <math>b=7</math> is a solution.
So <math>a-b=\boxed{\textbf{(C)}\ 3}</math>


== See Also ==
== See Also ==

Revision as of 11:38, 1 December 2013

Problem

Let $a$ and $b$ be relatively prime integers with $a>b>0$ and $\frac{a^3-b^3}{(a-b)^3}$ = $\frac{73}{3}$. What is $a-b$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution 1

Since $a$ and $b$ are both integers, so must $a^3-b^3$ and $(a-b)^3$. For this fraction to simplify to $\frac{73}{3}$, the denominator, or $a-b$, must be a multiple of 3. Looking at the answer choices, it is only possible when $a-b=\boxed{\textbf{(C)}\ 3}$.

Solution 2

Using difference of cubes in the numerator and cancelling out one $(a-b)$ in the numerator and denominator gives $\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}$.

Set $x = a^2 + b^2$, and $y = ab$. Then $\frac{x + y}{x - 2y} = \frac{73}{3}$. Cross multiplying gives $3x + 3y = 73x - 146y$, and simplifying gives $\frac{x}{y} = \frac{149}{70}$. Since $149$ and $70$ are relatively prime, we let $x = 149$ and $y = 70$, giving $a^2 + b^2 = 149$ and $ab = 70$. Since $a>b>0$, the only solution is $(a,b) = (10, 7)$, which can be seen upon squaring and summing the various factor pairs of $70$.


An alternate method of solving the system of equations involves solving the second equation for $a$, plugging it into the first equation, and solving the resulting quartic equation with a substitution of $u = b^2$. The four solutions correspond to $(\pm10, \pm7), (\pm7, \pm10)$


Thus, the desired quantity $a - b = \boxed{\textbf{(C)}\ 3}$.

Note that if you double $x$ and double $y$, you will get different (but not relatively prime) values for $a$ and $b$ that satisfy the original equation.

Solution 3

The first step is the same as above which gives $\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}$.

Then we can subtract $3ab$ and then add $3ab$ to get $\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}$, which gives $1+\frac{3ab}{(a-b)^2}=\frac{73}{3}$. $\frac{3ab}{(a-b)^2}=\frac{70}{3}$. Cross multiply $9ab=70(a-b)^2$. Since $a>b$, take the square root. $a-b=3\sqrt{\frac{ab}{70}}$. Since $a$ and $b$ are integers and relatively prime, $\sqrt{\frac{ab}{70}}$ is an integer. $ab$ is a multiple of $70$, so $a-b$ is a multiple of $3$. Therefore $a=10$ and $b=7$ is a solution. So $a-b=\boxed{\textbf{(C)}\ 3}$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_17&oldid=58226"