1987 AIME Problems/Problem 8

Problem

What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$ ?

Solution

Solution 1 Multiplying out all of the denominators, we get:

$\begin{align*}104(n+k) &< 195n< 105(n+k)\\ 0 &< 91n - 104k < n + k\end{align*}$

Since $91n - 104k < n + k$ , $k > \frac{6}{7}n$ . Also, $0 < 91n - 104k$ , so $k < \frac{7n}{8}$ . Thus, $48n < 56k < 49n$ . $k$ is unique if it is within a maximum range of $112$ , so $n = 112$ .

Solution 2 Flip all of the fractions for

$\frac{15}{8} > \frac{k + n}{n} > \frac{13}{7}$

$105n > 56 (k + n) > 104n$

$49n > 56k > 48n$

Continue as in Solution 1.

1987 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1987 AIME Problems/Problem 8

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