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 positive integers with <math>a>b>0</math> and <math>\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3} | + | Let <math>a</math> and <math>b</math> be relatively prime positive integers with <math>a>b>0</math> and <math>\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{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 6== | ==Solution 6== | ||
− | Let us rewrite the expression as <math>\frac{(a-b)^2 + 3ab | + | Let us rewrite the expression as <math>\frac{(a-b)^2 + 3ab}{(a-b)^2}</math>. Now letting <math>x = a - b</math>, we simplify the expression to <math>\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}</math>. Cross multiplying and doing a bit of simplification, we obtain that <math>ab = \frac{70x^2}{9}</math>. Since <math>a</math> and <math>b</math> are both integers, we know that <math>\frac{70x^2}{9}</math> has to be an integer. Experimenting with values of <math>x</math>, we get that <math>x = 3</math> which means <math>ab = 70</math>. We could prime factor from here to figure out possible values of <math>a</math> and <math>b</math>, but it is quite obvious that <math>a = 10</math> and <math>b=7</math>, so our desired answer is <math>\boxed{\textbf{(C)}\ 3}</math> ~triggod |
− | ~triggod | ||
Latest revision as of 17:04, 22 February 2021
Contents
Problem
Let and be relatively prime positive integers with and . What is ?
Solution 1
Since and are relatively prime, 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 .
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
Note:
From , the Euclidean Algorithm gives . Thus is relatively prime to , and clearly and are coprime as well. The solution must therefore be and .
Solution 4
Slightly expanding, we have that .
Canceling the , cross multiplying, and simplifying, we obtain that
. Dividing everything by , we get that
.
Applying the quadratic formula....and following the restriction that ....
.
Hence, .
Since they are relatively prime, , .
.
Solution 5
Note that the denominator, when simplified, gets We now have to test the answer choices. If one has a good eye or by simply testing the answer choices the answer will be clearly ~mathboy282
Solution 6
Let us rewrite the expression as . Now letting , we simplify the expression to . Cross multiplying and doing a bit of simplification, we obtain that . Since and are both integers, we know that has to be an integer. Experimenting with values of , we get that which means . We could prime factor from here to figure out possible values of and , but it is quite obvious that and , so our desired answer is ~triggod
Video Solution
https://youtu.be/ZWqHxc0i7ro?t=417
~ pi_is_3.14
Video Solution
~savannahsolver
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.