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 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>. | ||
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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 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 be seen upon squaring and summing the various factor pairs of .
An alternate method of solving the system of equations involves solving the second equation for , plugging it into the first equation, and solving the resulting quartic equation with a substitution of . The four solutions correspond to
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.
Solution 3
The first step is the same as above which gives .
Then we can subtract and then add to get , which gives . . Cross multiply . Since , take the square root. . Since and are integers and relatively prime, is an integer. is a multiple of , so is a multiple of . Therefore and is a solution. So
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.