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) | ||
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