Difference between revisions of "2007 AMC 10B Problems/Problem 25"
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+ | Also can refer to the same problem [https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_24| 2007 AMC 12B #24] (same problem) | ||
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=== Solution 1 === | === Solution 1 === | ||
Revision as of 00:35, 25 December 2020
Problem
How many pairs of positive integers are there such that and have no common factors greater than and:
is an integer?
Solution
Also can refer to the same problem 2007 AMC 12B #24 (same problem)
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 |
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