2018 AMC 12B Problems/Problem 17

Revision as of 22:37, 8 April 2019 by Olutosinfires (talk | contribs) (Solution 8)


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


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 $q-p=p+k-p=k\implies k\ge{\boxed{7}}$.

Solution 9

Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill. The interval can also be written as $0.5556<x<0.5714$. This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range.

The denominators to be considered are $9,10,11,12...$. We check $\frac{6}{10}, \frac{6}{11}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}$. At this point we know that we've got our fraction and our answer is $16-9=\boxed{textbf{A } 7}$

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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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