Find the smallest integer $n$ such that $\frac{11}{15}>\frac{m}{n}>\frac{7}{10}$ and $m$ is an integer.
There are multiple ways to solve this problem. We can get rid of the fractions by multiplying by $30n$ and do casework, we can convert to decimals and do casework, heck, we can graph and do casework. I know that this problem is really simple, but I am just looking for a way to solve this problem without casework.
It is clear that $m<n<2m$. So we can write $n=m+r$ with $0 < r <m$. Then we have $$\frac{7}{10} < \frac{m}{m+r}<\frac{11}{15}.$$ This is same as saying $$\frac{7r}{3} < m<\frac{11r}{4}.$$ Now we want the smallest $r$ (so that $n$ is smallest) for which $m$ is an integer. With $r=1$, the interval $\left(\frac{7}{3},\frac{11}{4}\right)$ will not contain any integer. With $r=2$ it does because $5 \in \left(\frac{14}{3},\frac{22}{4}\right)$, so $m=5$ and correspondingly $n=7$ is the smallest.