Difference between revisions of "2007 AMC 12B Problems/Problem 24"
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Revision as of 10:52, 4 July 2013
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
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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