Difference between revisions of "2007 AMC 12B Problems/Problem 24"
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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> | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | Let's assume that <math>\frac{a}{b} + \frac{14b}{9a} = m}</math> We get-- | ||
+ | |||
+ | <math>9a^2 - 9mab + 14b^2 = 0</math> | ||
+ | |||
+ | Factoring this, we get <math>4</math> equations- | ||
+ | |||
+ | <math>(3a-2b)(3a-7b) = 0</math> | ||
+ | |||
+ | <math>(3a-b)(3a-14b) = 0</math> | ||
+ | |||
+ | <math>(a-2b)(9a-7b) = 0</math> | ||
+ | |||
+ | <math>(a-b)(9a-14b) = 0</math> | ||
+ | |||
+ | (It's all subtractions, because if we had addition signs, that means <math>a</math> is the opposite sign of <math>b</math>) | ||
+ | |||
+ | Now we look at these, and see that- | ||
+ | |||
+ | <math>3a=2b</math> | ||
+ | |||
+ | <math>3a=b</math> | ||
+ | |||
+ | <math>3a=7b</math> | ||
+ | |||
+ | <math>3a=14b</math> | ||
+ | |||
+ | <math>a=2b</math> | ||
+ | |||
+ | <math>9a=7b</math> | ||
+ | |||
+ | <math>a=b</math> | ||
+ | |||
+ | <math>9a=14b</math> | ||
+ | |||
+ | This gives us <math>8</math> solutions, but we note that the middle term needs to give you back <math>9m</math>. | ||
+ | |||
+ | For example, in the case | ||
+ | |||
+ | <math>(a-2b)(9a-7b)</math>, the middle term is <math>-25ab</math>, which is not divisible by <math>9m</math> for whatever integar <math>m</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> | ||
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:19, 16 November 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 ,
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 subtractions, because if we had addition signs, that means is 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 divisible 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.