17. Let $p$ and $q$ be positive integers such that $\frac{72}{487}<\frac{p}{q}<\frac{18}{121}$. Find the smallest possible value of $q$.
(2 marks)
設 $p \vee q$ 為滿足 $\frac{72}{487}<\frac{p}{q}<\frac{18}{121}$ 的正整數。求 $q$ 的最小可能值。
(2 分)
17. 27
17. Taking reciprocals, we have $6 \frac{55}{72}=\frac{487}{72}>\frac{q}{p}>\frac{121}{18}=6 \frac{13}{18}$. Hence $q=6 p+r$ for some $0\frac{r}{p}>\frac{13}{18}$ and hence $\frac{72}{55} r<p<\frac{18}{13} r$.
To minimise $q$ we should therefore look for the smallest $r$ for which there exists an integer $p$ between $\frac{72}{55} r$ and $\frac{18}{13} r$. It is easy to check that the smallest such $r$ is 3 . This corresponds to $p=4$ and $q=6 \times 4+3=27$.