Let $a=bq+r$ with $0\le r\le b-1$, then $\frac{a}{b}=q+\frac{r}{b}>\frac{31}{17}$ then $q>\frac{16}{17}$ so $q\ge 1$. If $q=1$ then $\frac{r}{b}>\frac{16}{17}$ which implies $1>\frac{b}{17}>b-r>0$ (contradiction) So $q=2$ and $r=0$ (because $\frac{a}{b}$ in the least amount), then $\frac{a}{b}=2$

nmd27082001 wrote: ...then $\frac{a}{b}=2$ No : $\frac ab=\frac{11}6$ is better

pco wrote: nmd27082001 wrote: ...then $\frac{a}{b}=2$ No : $\frac ab=\frac{11}6$ is better You are right.

The problem with the first solution is saying that $31-17=16...$

That is wrong.

Write $\dfrac{a}{b}=2-\dfrac{c}{b}$ with a positive integer $c$. Then the condition $\dfrac{31}{17}<\dfrac{a}{b}$ translates into $\dfrac{3}{17}>\dfrac{c}{b}$. In order to minimize $\dfrac{a}{b}$, we need to maximize $\dfrac{c}{b}$. * Now $c\ge3$ (together with $b<17$) yields the contraditcion $\dfrac{c}{b}>\dfrac{3}{17}$. * For $c=2$ we get $b>\dfrac{2\cdot17}{3}=11+\dfrac13$. Hence $b\ge12$, and $\dfrac{c}{b}\ge\dfrac{2}{12}$ in this case. * For $c=1$ we get $b>\dfrac{17}{3}=5+\dfrac23$. Hence $b\ge6$, and $\dfrac{c}{b}\ge\dfrac{1}{6}$ in this case. Summarizing, the largest possible value for $\dfrac{c}{b}$ is $\dfrac{1}{6}$, and hence the smallest possible value for $\dfrac{a}{b}$ is $\dfrac{11}{6}$.