1. **State the problem:** We need to find the smallest positive integer $n$ such that there exists a positive integer $k$ where the inequality $$\frac{7}{13} < \frac{n}{n+k} < \frac{6}{11}$$ holds. 2. **Analyze the inequality:** The fraction $$\frac{n}{n+k}$$ is between two fractions, $$\frac{7}{13}$$ and $$\frac{6}{11}$$. 3. Since $n$ and $k$ are positive integers, rewrite the inequality: $$\frac{7}{13} < \frac{n}{n+k} < \frac{6}{11}$$ Multiply all parts by $n + k > 0$ to avoid inequality flip: $$\frac{7}{13} (n + k) < n < \frac{6}{11} (n + k)$$ 4. Rearrange each part to isolate $k$: From the left inequality: $$\frac{7}{13}(n + k) < n$$ $$\Rightarrow \frac{7}{13}n + \frac{7}{13}k < n$$ $$\Rightarrow \frac{7}{13}k < n - \frac{7}{13}n = n\left(1 - \frac{7}{13}\right) = n\frac{6}{13}$$ $$\Rightarrow k < \frac{6}{13} n \cdot \frac{13}{7} = \frac{6}{7} n$$ 5. From the right inequality: $$n < \frac{6}{11}(n + k)$$ $$\Rightarrow n < \frac{6}{11}n + \frac{6}{11}k$$ $$\Rightarrow n - \frac{6}{11}n < \frac{6}{11}k$$ $$\Rightarrow n\left(1 - \frac{6}{11}\right) < \frac{6}{11}k$$ $$\Rightarrow n \frac{5}{11} < \frac{6}{11}k$$ $$\Rightarrow \frac{5}{11} n < \frac{6}{11} k$$ $$\Rightarrow \frac{5}{6} n < k$$ 6. Combine inequalities for $k$: $$\frac{5}{6} n < k < \frac{6}{7} n$$ Since $k$ must be an integer, there must be at least one integer $k$ strictly between $\frac{5}{6} n$ and $\frac{6}{7} n$. 7. Check integers starting from the smallest $n$ given in the options: - For $n=9$: $$\frac{5}{6} \times 9 = 7.5$$ $$\frac{6}{7} \times 9 \approx 7.71$$ No integer $k$ satisfies $7.5 < k < 7.71$. - For $n=10$: $$\frac{5}{6} \times 10 = 8.33$$ $$\frac{6}{7} \times 10 \approx 8.57$$ No integer $k$ satisfies $8.33 < k < 8.57$. - For $n=11$: $$\frac{5}{6} \times 11 \approx 9.17$$ $$\frac{6}{7} \times 11 \approx 9.43$$ No integer $k$ satisfies $9.17 < k < 9.43$. - For $n=12$: $$\frac{5}{6} \times 12 = 10$$ $$\frac{6}{7} \times 12 \approx 10.29$$ No integer satisfies $10 < k < 10.29$. - For $n=13$: $$\frac{5}{6} \times 13 \approx 10.83$$ $$\frac{6}{7} \times 13 \approx 11.14$$ Integer $k=11$ satisfies $10.83 < 11 < 11.14$. 8. Verify that for $n=13$ and $k=11$ the original inequality holds: $$\frac{n}{n+k} = \frac{13}{13 + 11} = \frac{13}{24} \approx 0.5417$$ $$\frac{7}{13} \approx 0.5385 < 0.5417 < 0.5455 \approx \frac{6}{11}$$ So, $n=13$ works. **Final answer:** The smallest $n$ is 13. \boxed{13}