Difference between revisions of "2007 AMC 12B Problems/Problem 24"
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Since the denominator contains a factor of <math>9</math>, <math>9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b</math> | Since the denominator contains a factor of <math>9</math>, <math>9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b</math> | ||
− | + | Since <math>b = 3n</math> for some positive integer <math>n</math>, we can rewrite the fraction(divide by <math>9</math> on both top and bottom) as <math>\frac{a^2 + 14n^2}{3an}</math> | |
Since the denominator now contains a factor of <math>n</math>, we get <math>n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2</math>. | Since the denominator now contains a factor of <math>n</math>, we get <math>n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2</math>. | ||
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Thus there are four solutions: <math>(1,3)</math>, <math>(2,3)</math>, <math>(7,3)</math>, <math>(14,3)</math> and the answer is <math>\mathrm {(A)}</math> | Thus there are four solutions: <math>(1,3)</math>, <math>(2,3)</math>, <math>(7,3)</math>, <math>(14,3)</math> and the answer is <math>\mathrm {(A)}</math> | ||
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==Solution 2== | ==Solution 2== |
Revision as of 10:57, 25 December 2013
Contents
[hide]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 ,
Since for some positive integer , we can rewrite the fraction(divide by on both top and bottom) 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.