Difference between revisions of "2007 AMC 12B Problems/Problem 24"
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Clearing the denominator and simplifying, we get a quadratic in terms of <math>u</math>: | Clearing the denominator and simplifying, we get a quadratic in terms of <math>u</math>: | ||
− | <math>9u^2 - 9au + 14 = 0 \Rightarrow u = \frac{9a \pm \sqrt{(9a^2 | + | <math>9u^2 - 9au + 14 = 0 \Rightarrow u = \frac{9a \pm \sqrt{(9a)^2 - 504}}{18}</math> |
− | Since <math>a</math> and <math>b</math> | + | Since <math>a</math> and <math>b</math> are integers, <math>u</math> is a rational number. This means that <math>\sqrt{(9a)^2 - 504}</math> is an integer. |
+ | |||
+ | Let <math>\sqrt{(9a)^2 - 504} = x</math>. Squaring and rearranging yields: | ||
+ | |||
+ | <math>(9a)^2 - x^2 = 504</math> | ||
+ | |||
+ | <math>(9a+x)(9a-x) = 504</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 14:16, 29 November 2014
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 :
Since and
are integers,
is a rational number. This means that
is an integer.
Let . Squaring and rearranging yields:
.
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 |
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