Let p and q be positive integers such that (5)/(9)<(p)/(q)<(4)/(7) and q is as small as possible. What is q-p?

2018 AMC 12B Problem 17

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17.

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

Since

9p\gt5q

and

4q\gt7p

with integer sides,

9p-5q\ge1

and

4q-7p\ge1

Add these after writing

\tfrac{4}{7}-\tfrac{5}{9}=\tfrac{1}{63}

to bound

q

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,

p=9

and

q-p=16-9=7.

Thus, the correct answer is A.