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

Revision as of 14:43, 11 July 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 $(16,9)$ 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 using vectors.

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 $\boxed{\textbf{(A)}\ 7}$ . 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 $\frac{4}{7}\approx 0.6$

Solution 6

Inverting the given inequality we get $\frac{7}{4} < \frac{q}{p} < \frac{9}{5}$

which simplifies to $35p < 20q < 36p$

We can now substitute $q = p + k$ . Note we need to find $k$ .

$35p < 20p + 20k < 36p$

which simplifies to $15p < 20k < 16p$

Cleary $p$ is greater than $k$ . We will now substitute $p = k + x$ to get

$15k + 15x < 20k < 16k + 16x$

The inequality $15k + 15x < 20k$ simplifies to $3x < k$ . The inequality $20k < 16k + 16x$ simplifies to $k < 4x$ . Combining the two we get $3x < k < 4x$

Since $x$ and $k$ are integers, the smallest values of $x$ and $k$ that satisfy the above equation are $2$ and $7$ respectively. Substituting these back in, we arrive with an answer of $\boxed{\textbf{(A)}\ 7}$ .

Solution 7

Start with $\frac{5}{9}$ . Repeat the following process until you arrive at the answer: if the fraction is less than or equal to $\frac{5}{9}$ , add $1$ to the numerator; otherwise, if it is greater than or equal to $\frac{4}{7}$ , add one $1$ to the denominator. We have:

$\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}$

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

Solution 8

Because q and p are positive integers with $p<q$ , we can let $q=p+k$ where $k\in{\mathbb{Z}}$ . Now, the problem condition reduces to

$\frac{5}{9}<\frac{p}{p+k}<\frac{4}{7}$

Our first inequality is $\frac{5}{9}<\frac{p}{p+k}$ which gives us $5p+5k<9p\implies \frac{5}{4}k<p$ .

Our second inequality is $\frac{p}{p+k}<\frac{4}{7}$ which gives us $7p<4p+4k\implies p<\frac{4}{3}k$ .

Hence, $\frac{5}{4}k<p<\frac{4}{3}k\implies 15k<12p<16k$ .

It is clear that we are aiming to find the least positive integer value of k such that there is at least one value of p that satisfies the inequality.

Now, simple casework through the answer choices of the problem reveals that q-p=p+k-p=k\implies k\ge{\boxed{7}}$.

@@ Line 81: / Line 81: @@
 Our first inequality is <math>\frac{5}{9}<\frac{p}{p+k}</math> which gives us <math>5p+5k<9p\implies \frac{5}{4}k<p</math>.
-Our second inequality is <math>\frac{p}{p+k}<\frac{4}{7} which gives us </math>7p<4p+4k\implies p<\frac{4}{3}k<math>.
+Our second inequality is <math>\frac{p}{p+k}<\frac{4}{7}</math> which gives us <math>7p<4p+4k\implies p<\frac{4}{3}k</math>.
-Hence, </math>\frac{5}{4}k<p<\frac{4}{3}k\implies 15k<12p<16k<math>.
+Hence, <math>\frac{5}{4}k<p<\frac{4}{3}k\implies 15k<12p<16k</math>.
 It is clear that we are aiming to find the least positive integer value of k such that there is at least one value of p that satisfies the inequality.
-Now, simple casework through the answer choices of the problem reveals that q-p=p+k-p=k\implies k\ge{\boxed{7}}</math>.
+Now, simple casework through the answer choices of the problem reveals that q-p=p+k-p=k\implies k\ge{\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
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