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2018 AMC 12B Problem 17

Problem 17 of 25IntermediateAlgebra

Let pp and qq be positive integers such that 59<pq<47 \dfrac{5}{9}\lt\dfrac{p}{q}\lt\dfrac{4}{7} and qq is as small as possible. What is q−p?q-p?

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Solution

From 59<pq\tfrac59\lt\tfrac pq we get 9p−5q≥1,9p-5q\ge1, and from pq<47\tfrac pq\lt\tfrac47 we get 4q−7p≥1.4q-7p\ge1. Now 163=47−59=4q−7p7q+9p−5q9q≥17q+19q=1663q. \begin{gathered} \dfrac{1}{63}=\dfrac47-\dfrac59 \\ =\dfrac{4q-7p}{7q}+\dfrac{9p-5q}{9q} \\ \ge\dfrac{1}{7q}+\dfrac{1}{9q} \\ =\dfrac{16}{63q}. \end{gathered} Hence q≥16.q\ge16. With q=16,q=16, the fraction 916\tfrac{9}{16} lies strictly between 59\tfrac59 and 47,\tfrac47, so p=9p=9 and q−p=16−9=7.q-p=16-9=7. Thus, the correct answer is A.

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Concepts: fraction · inequality · bounding to limit cases

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.