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>\ | + | 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> |
− | == Solution 1== | + | ==Solution 1== |
− | Since <math>a</math> and <math>b</math> are | + | Since <math>a</math> and <math>b</math> are relatively prime, <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>. |
− | == 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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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 be seen upon squaring and summing the various factor pairs of <math>70</math>. | 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 be seen upon squaring and summing the various factor pairs of <math>70</math>. | ||
+ | Thus, <math>a - b = \boxed{\textbf{(C)}\ 3}</math>. | ||
− | + | '''Remarks:''' | |
+ | An alternate method of solving the system of equations involves solving the second equation for <math>a</math>, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of <math>u = b^2</math>. The four solutions correspond to <math>(\pm10, \pm7), (\pm7, \pm10).</math> | ||
− | + | Also, we can solve for <math>a-b</math> directly instead of solving for <math>a</math> and <math>b</math>: <math>a^2-2ab+b^2=149-2(70)=9 \implies a-b=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. | 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> | ||
+ | |||
+ | '''Note:''' | ||
+ | |||
+ | From <math>9ab=70(a-b)^2</math>, the Euclidean Algorithm gives <math>\gcd(a-b,a)=\gcd(a-b,b)=1</math>. Thus <math>(a-b)^2</math> is relatively prime to <math>ab</math>, and clearly <math>9</math> and <math>70</math> are coprime as well. The solution must therefore be <math>(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}</math> and <math>ab=70</math>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | Slightly expanding, we have that | ||
+ | <math>\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}</math>. | ||
+ | |||
+ | Canceling the <math>(a-b)</math>, cross multiplying, and simplifying, we obtain that | ||
+ | |||
+ | <math>0=70a^2-149ab+70b^2</math>. | ||
+ | Dividing everything by <math>b^2</math>, we get that | ||
+ | |||
+ | <math>0=70(\frac{a}{b})^2-149(\frac{a}{b})+70</math>. | ||
+ | |||
+ | Applying the quadratic formula....and following the restriction that <math>a>b>0</math>.... | ||
+ | |||
+ | <math>\frac{a}{b}=\frac{10}{7}</math>. | ||
+ | |||
+ | Hence, <math>7a=10b</math>. | ||
+ | |||
+ | Since they are relatively prime, <math>a=10</math>, <math>b=7</math>. | ||
+ | |||
+ | <math>10-7=\boxed{\textbf{(C)}\ 3}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Note that the denominator, when simplified, gets <math>3.</math> 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 <math>\boxed{\textbf{(C)}\ 3}</math> ~mathboy282 | ||
+ | |||
+ | |||
+ | ==Solution 6== | ||
+ | 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 | ||
+ | |||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/ZWqHxc0i7ro?t=417 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/8SXVrlH71jk | ||
+ | |||
+ | ~savannahsolver | ||
== 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:33, 20 January 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.