Difference between revisions of "2007 AMC 12B Problems/Problem 24"
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How many pairs of positive integers <math>(a,b)</math> are there such that <math>\gcd(a,b)=1</math> and <cmath>\frac{a}{b}+\frac{14b}{9a}</cmath> is an integer? | How many pairs of positive integers <math>(a,b)</math> are there such that <math>\gcd(a,b)=1</math> and <cmath>\frac{a}{b}+\frac{14b}{9a}</cmath> is an integer? | ||
− | <math>\mathrm {(A)} 4</math> | + | <math>\mathrm {(A)} 4</math> <math>\mathrm {(B)} 6</math> <math>\mathrm {(C)} 9</math> <math>\mathrm {(D)} 12</math> <math>\mathrm {(E)} \text{infinitely many}</math> |
==Solution== | ==Solution== |
Revision as of 17:52, 24 December 2013
Problem 24
How many pairs of positive integers are there such that and is an integer?
Solution
Combining the fraction, must be an integer.
Since the denominator contains a factor of ,
Rewriting as for some positive integer , we can rewrite the fraction as
Since the denominator now contains a factor of , we get .
But since , we must have , and thus .
For the original fraction simplifies to .
For that to be an integer, must divide , and therefore we must have . Each of these values does indeed yield an integer.
Thus there are four solutions: , , , and the answer is
Solution 2
Let's assume that $\frac{a}{b} + \frac{14b}{9a} = m}$ (Error compiling LaTeX. Unknown error_msg) We get--
Factoring this, we get equations-
(It's all negative, because if we had positive signs, would be the opposite sign of )
Now we look at these, and see that-
This gives us solutions, but we note that the middle term needs to give you back .
For example, in the case
, the middle term is , which is not equal by for whatever integar .
Similar reason for the fourth equation. This elimnates the last four solutions out of the above eight listed, giving us 4 solutions total
Solution 3
Beat it! We get the awesome answer of A!
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Solution 3: We are awesome(duh). We are awesome at guessing and fail and guess E. There are zero solutions. LOLLOL