Mmmmm

You could use the fact that if you have a fraction less than 1 then: $\frac{m}{n} < \frac{m+1}{n+1}$ (which can be seen by croos multiplying to obtain m < n ). Now appling this to both sides: $\frac{8}{11}\leq\frac{m}{n}\leq\frac{5}{7}$ However, this step cannot be repeated otherwise we obtain $\frac{3}{4}\leq\frac{m}{n}\leq\frac{2}{3}$ $\therefore\ min(n) = 7$

For $n=1$

$0<\frac{7}{10}<\frac{m}{n}<\frac{11}{15}<1$

For $n=2$

$\frac{1}{2}=\frac{5}{10}<\frac{7}{10}<\frac{m}{n}<\frac{11}{15}<\frac{2}{2}$

For $n=3$

$\frac{2}{3}<\frac{7}{10}<\frac{m}{n}<\frac{11}{15}<\frac{3}{3}$

For $n=4$

$\frac{2}{4}<\frac{7}{10}<\frac{m}{n}<\frac{11}{15}<\frac{3}{4}$

For $n=5$

$\frac{3}{5}<\frac{7}{10}<\frac{m}{n}<\frac{11}{15}<\frac{4}{5}$

For $n=6$

$\frac{4}{6}<\frac{7}{10}<\frac{m}{n}<\frac{11}{15}<\frac{5}{6}$

Finally, for $n=7$

$\frac{49}{70}<\frac{50}{70}$ and $\frac{50}{70}=\frac{75}{105}<\frac{77}{105}$

The minimum value for $n$ is $\boxed{n=7}$

Multiply all expressions by 30n . The inequality remains valid as 30n is positive. It now becomes: $21n<30m<22n$ $\therefore 21n<30m<21n+n$ This simply means that we have to find the first multiple of 30 that occurs between 21n and 21n+n , for some smallest integer n . For that, we just have to analyze the table of 21 : $21\leftrightarrow 22$ $42\leftrightarrow 44$ $63\leftrightarrow 66$ $84\leftrightarrow 88$ $105\leftrightarrow 110$ $126\leftrightarrow 132$ $147\leftrightarrow 154$ $150=5\times 30$ There! 150 , a multiple of 30 lies between the 7th multiples of 21 and 22 . Thus, we have m=5 and n=7 .