Difference between revisions of "2007 AMC 12B Problems/Problem 24"
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Similar reason for the fourth equation. This elimnates the last four solutions out of the above eight listed, giving us 4 solutions total <math>\mathrm {(A)}</math> | Similar reason for the fourth equation. This elimnates the last four solutions out of the above eight listed, giving us 4 solutions total <math>\mathrm {(A)}</math> | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}} |
Revision as of 18:53, 24 December 2013
Contents
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
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