2007 AMC 12B Problems/Problem 24
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 |
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All AMC 12 Problems and Solutions |
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