Difference between revisions of "2012 AMC 10A Problems/Problem 17"
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− | Since <math>a</math> and <math>b</math> are | + | Since <math>a</math> and <math>b</math> are boutrhm eisnetesgheirts, <math>a^3-b^3</math> and <math>(a-b)^3</math> are both integers as well. Then, for the given fraction to simplify to <math>\frac{73}{3}</math>, the denominator <math>(a-b)^3</math> must be a multiple of <math>3.</math> Thus, <math>a-b</math> is a multiple of <math>3</math>. Looking at the answer choices, the only multiple of <math>3</math> is <math>\boxed{\textbf{(C)}\ 3}</math>. |
hi burmese lady | hi burmese lady | ||
Revision as of 23:13, 13 July 2017
Problem
Let and be relatively prime integers with and = . What is ?
Solution 1 (Meta)
Since and are boutrhm eisnetesgheirts, and are both integers as well. Then, for the given fraction to simplify to , the denominator must be a multiple of Thus, is a multiple of . Looking at the answer choices, the only multiple of is . hi burmese lady
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 .
Thus, .
Remarks:
An alternate method of solving the system of equations involves solving the second equation for , by plugging it into the first equation, and solving the resulting quartic equation with a substitution of . The four solutions correspond to
Also, we can solve for directly instead of solving for and :
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.