2007 AMC 12B Problems/Problem 24
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
Solution 3
Let . Then the given equation becomes
.
Let's set this equal to some value, .
Clearing the denominator and simplifying, we get a quadratic in terms of :
$9u^2 - 9au + 14 = 0 \Rightarrow u = \frac{9a \pm \sqrt{(9a)^2 - 4\cdot 9 \cdot 14}{18}$ (Error compiling LaTeX. Unknown error_msg)
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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All AMC 12 Problems and Solutions |
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