Difference between revisions of "2018 AMC 12B Problems/Problem 17"

Revision as of 00:58, 19 February 2018

Problem

Let $p$ and $q$ be positive integers such that $\frac{5}{9} < \frac{p}{q} < \frac{4}{7}$ and $q$ is 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 $\boxed{\textbf{(A)}\ 7}$ .

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 $\boxed{\textbf{(A)}\ 7}$

Solution 3

Cross-multiply the inequality to get $35q < 63p < 36q.$

Then, $0 < 63p-35q < q,$ $0 < 7(9p-5q) < q.$

Since $p$ , $q$ are integers, $9p-5q$ is an integer. To minimize $q$ , start from $9p-5q=1$ , which gives $p=\frac{5q+1}{9}$ . This limits $q$ to be greater than $7$ , so test values of $q$ starting from $q=8$ . However, $q=8$ to $q=14$ do not give integer values of $p$ .

Once $q>14$ , it is possible for $9p-5q$ to be equal to $2$ , so $p$ could also be equal to $\frac{5q+2}{9}.$ The next value, $q=15$ , is not a solution, but $q=16$ gives $p=\frac{5\cdot 16 + 1}{9} = 9$ . Thus, the smallest possible value of $q$ is $16$ , and the answer is $16-9= \boxed{\textbf{(A)}\ 7}$ .

Solution 4

Graph the regions $y > \frac{5}{9}x$ and $y < \frac{4}{7}x$ . Note that the lattice point (9,16) is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is $\frac{9}{16}$ and the answer is $16-9= \boxed{\textbf{(A)}\ 7}$ .

Remark: This also gives an intuitive geometric proof of the mediant.

Solution 5 (Using answer choices to prove mediant)

As the other solutions do, the mediant $=\frac{9}{16}$ is between the two fractions, with a difference of $\boxed{(A) 7}. Suppose that the answer was not$ A $, then the answer must be$ B $or$ C $as otherwise$ p $would be negative. Then, the possible fractions with lower denominator would be$ \frac{k-11}{k} $for$ k=12,13,14,15 $and$ \frac{k-13}{k} $for$ k=14,15, $which are clearly not anywhere close to$ \frac{4}{7}\approx 0.6$

@@ Line 39: / Line 39: @@
 Remark: This also gives an intuitive geometric proof of the mediant.
+==Solution 5 (Using answer choices to prove mediant)==
+As the other solutions do, the mediant <math>=\frac{9}{16}</math> is between the two fractions, with a difference of <math>\boxed{(A) 7}. Suppose that the answer was not </math>A<math>, then the answer must be </math>B<math> or </math>C<math> as otherwise </math>p<math> would be negative. Then, the possible fractions with lower denominator would be </math>\frac{k-11}{k}<math> for </math>k=12,13,14,15<math> and </math>\frac{k-13}{k}<math> for </math>k=14,15,<math> which are clearly not anywhere close to </math>\frac{4}{7}\approx 0.6$
 ==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
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