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>\frac{a^3-b^3}{(a-b)^3}</math> = <math>\frac{73}{3}</math>. What is <math>a-b</math>? | Let <math>a</math> and <math>b</math> be relatively prime integers with <math>a>b>0</math> and <math>\frac{a^3-b^3}{(a-b)^3}</math> = <math>\frac{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> | ||
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| + | == Solution == | ||
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| + | 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>. | ||
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| + | == See Also == | ||
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| + | {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}} | ||
Revision as of 00:13, 9 February 2012
Problem
Let 
and 
be relatively prime integers with 
and 
= 
. What is 
?
Solution
Since and are both integers, so must 
and 
. For this fraction to simplify to , the denominator, or , must be a multiple of 3. Looking at the answer choices, it is only possible when 
.
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 | ||
