Difference between revisions of "2007 AMC 12B Problems/Problem 24"
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<math>(a-b)(9a-14b) = 0</math> | <math>(a-b)(9a-14b) = 0</math> | ||
− | (It's all | + | (It's all negative, because if we had positive signs, <math>a</math> would be the opposite sign of <math>b</math>) |
Now we look at these, and see that- | Now we look at these, and see that- | ||
Line 65: | Line 65: | ||
For example, in the case | For example, in the case | ||
− | <math>(a-2b)(9a-7b)</math>, the middle term is <math>-25ab</math>, which is not | + | <math>(a-2b)(9a-7b)</math>, the middle term is <math>-25ab</math>, which is not equal 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> | 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> |
Revision as of 22:23, 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 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 |
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