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

Revision as of 17:28, 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 $\boxed{\textbf{(A)}\ 7}$ . (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 $\boxed{\textbf{(A)}\ 7}$

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 $16-9= \boxed{\textbf{(A)}\ 7}$ .

@@ Line 10: / Line 10: @@
 <cmath>\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},</cmath>
-which reduces to <math>q\geq b+d</math>. We can easily find that <math>p=a+c</math>, giving an answer of <math>16-9=\boxed{7}</math>. (pieater314159)
+which reduces to <math>q\geq b+d</math>. We can easily find that <math>p=a+c</math>, giving an answer of <math>\boxed{\textbf{(A)}\ 7}</math>. (pieater314159)
 ==Solution 2 (requires justification)==
@@ Line 22: / Line 22: @@
 <math>4q - 7p = 1</math>
-Solving the system of equations yields <math>q=16</math> and <math>p=9</math>. Therefore, the answer is <math>16-9=\boxed{7}</math>
+Solving the system of equations yields <math>q=16</math> and <math>p=9</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 7}</math>
+==Solution 3==
+Cross-multiply the inequality to get <cmath>35q < 63p < 36q.</cmath>
+Then,
+<cmath>0 < 63p-35q < q,</cmath>
+<cmath>0 < 7(9p-5q) < q.</cmath>
+Since <math>p</math>, <math>q</math> are integers, <math>9p-5q</math> is an integer. To minimize <math>q</math>, start from <math>9p-5q=1</math>, which gives <math>p=\frac{5q+1}{9}</math>. This limits <math>q</math> to be greater than <math>7</math>, so test values of <math>q</math> starting from <math>q=8</math>. However, <math>q=8</math> to <math>q=14</math> do not give integer values of <math>p</math>.
+Once <math>q>14</math>, it is possible for <math>9p-5q</math> to be equal to <math>2</math>, so <math>p</math> could also be equal to <math>\frac{5q+2}{9}.</math> The next value, <math>q=15</math>, is not a solution, but <math>q=16</math> gives <math>p=\frac{5\cdot 16 + 1}{9} = 9</math>. Thus, the smallest possible value of <math>q</math> is <math>16</math>, and the answer is <math>16-9= \boxed{\textbf{(A)}\ 7}</math>.
 ==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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