2007 AMC 10B Problems/Problem 25
Problem
How many pairs of positive integers are there such that
and
have no common factors greater than
and:
is an integer?
Solution
Solution 1
We will use the divisibility notation (), which means
is divisible by
. Getting common denominators, we have to find coprime
such that
.
is divisible by 3 because 14 is not a multiple of three in the equation, so
must balance it and make them integers. Since
and
are coprime,
. Similarly,
. However,
cannot be
as
only has solutions when
. Therefore,
and
. Checking them all (or noting that
is the smallest answer choice), we see that they work and the answer is
.
Solution 2
Let . We can then write the given expression as
where
is an integer. We can rewrite this as a quadratic,
. By the Quadratic Formula,
. We know that
must be rational, so
must be a perfect square. Let
. Then,
. The factors pairs of
are
and
,
and
,
and
, and
and
. Only
and
and
and
give integer solutions,
and
and
and
, respectively. Plugging these back into the original equation, we get
possibilities for
, namely
and
.
See Also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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