Difference between revisions of "2007 AMC 10B Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | How many pairs of positive integers (a,b) are there such that a and b have no common factors greater than 1 and: | + | How many pairs of positive integers <math>(a,b)</math> are there such that <math>a</math> and <math>b</math> have no common factors greater than <math>1</math> and: |
− | < | + | <cmath>\frac{a}{b} + \frac{14b}{9a}</cmath> |
is an integer? | is an integer? |
Revision as of 16:47, 18 September 2020
Contents
[hide]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 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.