Aug 3, 2020
AMC 10-2007 (B25)
How many pairs of positive integers (a, b) are there such that a and b have no common factors greater than 1 and a/b + 14b/a is an integer?
(A) 4 (B) 6 (C) 9 (D) 12 (E) infinitely many
We can rewrite the fractions as (9a^2 + 14b^2) / 9ab. Therefore, 3|b. Since a and b are coprime, a|9a^2 + 14b^2, so a|14. Similarly, b|9. However, b cannot be 9 as 81a|9a^2 + 81*14 only has solutions when 3|a, which can’t happen when a|14. So, b=3, and a can be {1, 2, 7, 14}. Checking shows us that all 4 work, so the answer is A (4).