2018 AMC 12B Problems/Problem 17

Problem

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

$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution 1

We claim that, between any two fractions $a/b$ and $c/d$ , if $bc-ad=1$ , the fraction with smallest denominator between them is $\frac{a+c}{b+d}$ . To prove this, we see that

$\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},$ which reduces to $q\geq b+d$ . We can easily find that $p=a+c$ , giving an answer of $16-9=\boxed{7}$ . (pieater314159)

Solution 2 (requires justification)

Assume that the difference $\frac{p}{q} - \frac{5}{9}$ results in a fraction of the form $\frac{1}{9q}$ . Then, $9p - 5q = 1$ Also assume that the difference $\frac{p}{q} - \frac{5}{9}$ results in a fraction of the form $\frac{1}{9q}$ . Then, $9p - 5q = 1$

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2018 AMC 12B Problems/Problem 17

Contents

Problem

Solution 1

Solution 2 (requires justification)

See Also