2012 AMC 10A Problems/Problem 17

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 relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\frac{73}{3}$ , the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$ . Looking at the answer choices, the only multiple of $3$ is $\boxed{\textbf{(C)}\ 3}$ .

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 $70$ .

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

Remarks:

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

Also, we can solve for $a-b$ directly instead of solving for $a$ and $b$ : $a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.$

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 $a-b=\boxed{\textbf{(C)}\ 3}$

Solution 4

Slightly expanding, we have that $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$ .

Canceling the $(a-b)$ , cross multiplying, and simplifying, we obtain that

$0=70a^2-149ab+70b^2$ . Dividing everything by $b^2$ , we get that

$0=70(\frac{a}{b})^2-149(\frac{a}{b})+70$ .

Applying the quadratic formula....and following the restriction that $a>b>0$ ....

$\frac{a}{b}=\frac{10}{7}$ .

Hence, $7a=10b$ .

Since they are relatively prime, $a=10$ , $b=7$ .

$10-7=\boxed{\textbf{(C)}\ 3}$ .

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
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