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

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

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2012 AMC 10A Problems/Problem 17

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