Difference between revisions of "2012 AMC 10A Problems/Problem 17"
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<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 == | + | == 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== | ||
+ | |||
+ | 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>. | ||
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+ | Set <math>x = a^2 + b^2</math>, and <math>y = ab</math>. Then <math>\frac{x + y}{x - 2y} = \frac{73}{3}</math>. Cross multiplying gives <math>3x + 3y = 73x - 146y</math>, and simplifying gives <math>\frac{x}{y} = \frac{149}{70}</math>. Since <math>149</math> and <math>70</math> are relatively prime, we let <math>x = 149</math> and <math>y = 70</math>, giving <math>a^2 + b^2 = 149</math> and <math>ab = 70</math>. Since <math>a>b>0</math>, the only solution is <math>(a,b) = (10, 7)</math>, which can either be seen upon squaring and summing the various factor pairs of <math>70</math>. | ||
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+ | Thus, the desired quantity <math>a - b = \boxed{\textbf{(C)}\ 3}</math>. | ||
+ | |||
+ | 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. | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}} |
Revision as of 00:50, 9 February 2012
Contents
Problem
Let and be relatively prime integers with and = . What is ?
Solution 1
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 .
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 either be seen upon squaring and summing the various factor pairs of .
Thus, the desired quantity .
Note that if you double and double , you will get different (but not relatively prime) values for and that satisfy the original equation.
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 |