Question

Let $p$ and $q$ be positive integers such that $\dfrac{5}{9}\lt\dfrac{p}{q}\lt\dfrac{4}{7}$ and $q$ is as small as possible. What is $q-p?$

$7$

$11$

$13$

$17$

$19$

Answer: A

Concepts:fraction inequality bounding to limit cases

Difficulty rating: 2090

Solution:

From $\tfrac59\lt\tfrac pq$ we get $9p-5q\ge1,$ and from $\tfrac pq\lt\tfrac47$ we get $4q-7p\ge1.$ Now $\begin{gathered} \dfrac{1}{63}=\dfrac47-\dfrac59 \\ =\dfrac{4q-7p}{7q}+\dfrac{9p-5q}{9q} \\ \ge\dfrac{1}{7q}+\dfrac{1}{9q} \\ =\dfrac{16}{63q}. \end{gathered}$

Hence $q\ge16.$ With $q=16,$ the fraction $\tfrac{9}{16}$ lies strictly between $\tfrac59$ and $\tfrac47,$ so $p=9$ and $q-p=16-9=7.$

Thus, the correct answer is A.

2018 AMC 12B Problem 17

Answer: A

Difficulty rating: 2090

Solution:

Problem 17 in Other Years