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

We can write the inequality as

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

Thus, multiplying the members of the inequality by $56n$, we obtain the equivalent inequality

$$
\begin{aligned}
104n & <56n+56k & <105n \\
48n< & 56k & <49n
\end{aligned}
$$

For there to be a unique integer $k$ satisfying the inequality, the interval $(48n, 49n)$ must contain exactly one multiple of 56. The length of such an interval is $n$ and it contains exactly $n-1$ positive integers. If $n-1 \geq 2 \cdot 56=112$, the interval will contain at least two multiples of 56 and will not work for the inequality. Therefore, $n \leq 112$. To verify that $n=112$ is the maximum solution, it is sufficient to note that

$$
\begin{aligned}
48 \cdot 112 & =56 \cdot 96 \\
& <56 \cdot 97 \\
& <56 \cdot 98 \\
& =49 \cdot 112
\end{aligned}
$$

Thus, $k=97$ is the only possible integer to satisfy the inequality when $n=112$.