The simplest fraction between one third and two fifths

Statement

In a class, more than $\dfrac{1}{3}$ but less than $\dfrac{2}{5}$ of the students cycle to school. What is the smallest number of students the class can have?

Solution

We seek the smallest $n$ for which there is an integer $k$ (the cyclists) with $\frac{1}{3} < \frac{k}{n} < \frac{2}{5}.$ A natural candidate is the mediant $\dfrac{1+2}{3+5}=\dfrac{3}{8}$, giving $n=8$, $k=3$: $\frac{1}{3}\approx 0.333 < \frac{3}{8}=0.375 < 0.4 = \frac{2}{5}.$ For every $n\le 7$ there is no integer strictly between $\dfrac{n}{3}$ and $\dfrac{2n}{5}$. Hence the minimum is $n = \boxed{8}.$