Difference between revisions of "2012 AMC 10A Problems/Problem 17"
(→Solution 4) |
|||
Line 39: | Line 39: | ||
Slightly expanding, we have that | Slightly expanding, we have that | ||
<math>\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}</math>. | <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 | Canceling the <math>(a-b)</math>, cross multiplying, and simplifying, we obtain that | ||
+ | |||
<math>0=70a^2-149ab+70b^2</math>. | <math>0=70a^2-149ab+70b^2</math>. | ||
Dividing everything by <math>b^2</math>, we get that | Dividing everything by <math>b^2</math>, we get that | ||
+ | |||
<math>0=70(\frac{a}{b})^2-149(\frac{a}{b})+70</math>. | <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> | + | |
+ | Applying the quadratic formula....and following the restriction that <math>a>b>0</math>.... | ||
+ | |||
+ | |||
<math>\frac{a}{b}=\frac{10}{7}</math>. | <math>\frac{a}{b}=\frac{10}{7}</math>. | ||
+ | |||
Hence, <math>7a=10b</math>. | Hence, <math>7a=10b</math>. | ||
+ | |||
Since they are relatively prime, <math>a=10</math>, <math>b=7</math>. | Since they are relatively prime, <math>a=10</math>, <math>b=7</math>. | ||
+ | |||
<math>10-7=\boxed{\textbf{(C)}\ 3}</math>. | <math>10-7=\boxed{\textbf{(C)}\ 3}</math>. | ||
+ | |||
== 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}} | {{MAA Notice}} |
Revision as of 15:58, 4 February 2018
Contents
Problem
Let and be relatively prime integers with and = . What is ?
Solution 1 (The not so good solution)
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
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, , .
.
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.