A problem on Farey Sequences

Let $n$ be a positive integer such that there exists $m \in \mathbb{N}$ coprime to $n$ for which $\frac{2}{3} < \frac{m}{n} < \frac{7}{10}$ . Then, consider the equivalent inequality $\dfrac{10}{7} m < n < \dfrac{3}{2} m$ .

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• Case 1: $\frac{3}{2} m - \frac{10}{7} m \ge 1$

  It follows $m \ge 14$ and, consequently, $n > 20$ .

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• Case 2: $\frac{3}{2} m - \frac{10}{7} m < 1$

  $\dfrac{3}{2} m$ and $\dfrac{10}{7} m$ cannot be integers (hence $m$ is odd and is not divisible by $7$ ), because they are separated by $n$ . For the same reason, we have $\begin{aligned} \Bigl \lceil \dfrac{10}{7} m \Bigr \rceil &= \Bigl \lfloor \dfrac{3}{2} m \Bigr \rfloor \\ \text{} \\ \Bigl \lceil \dfrac{10}{7} (2t-&1) \Bigr \rceil = 3t-2 & \text{if we let m=2t-1} \\ \text{} \\ 3t-3 < \dfrac{10}{7} (&2t-1) \le 3t-2 \\ \text{} \\ 4 \le &t <11 \end{aligned}$

Note that, in case 2 , minimizing $m$ is sufficient to find the smallest $n$ of that case. Now, if $t$ is $4$ , then $m$ would be divisible by $7$ . Hence, the minimum value of $t$ is $5$ , so that the minimum of $m$ is $9$ . And so, finally, $n=13$ is the smallest solution for case 2 .

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However, since this last value is less than $20$ , it is also the overall minimum, so that the answer is $\boxed{13}$ .