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Difference between revisions of "2018 AMC 12B Problems/Problem 17"
Line 15: Line 15:
  
 
Assume that the difference <math>\frac{p}{q} - \frac{5}{9}</math> results in a fraction of the form <math>\frac{1}{9q}</math>. Then,
 
Assume that the difference <math>\frac{p}{q} - \frac{5}{9}</math> results in a fraction of the form <math>\frac{1}{9q}</math>. Then,
−
 
 
<math>9p - 5q = 1</math>
 
<math>9p - 5q = 1</math>
  
 
Also assume that the difference <math>\frac{4}{7} - \frac{p}{q}</math> results in a fraction of the form <math>\frac{1}{7q}</math>. Then,
 
Also assume that the difference <math>\frac{4}{7} - \frac{p}{q}</math> results in a fraction of the form <math>\frac{1}{7q}</math>. Then,
−
 
 
<math>4q - 7p = 1</math>
 
<math>4q - 7p = 1</math>
  

Revision as of 16:53, 16 February 2018

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 $\frac{4}{7} - \frac{p}{q}$ results in a fraction of the form $\frac{1}{7q}$. Then, $4q - 7p = 1$

Solving the system of equations yields $q=16$ and $p=9$. Therefore, the answer is $16-9=\boxed{7}$

See Also

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

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