How find this minimum of the $q$, if such $\frac{95}{36}>\frac{p}{q}>\frac{96}{37}$

Question

let $p,q$ is postive integer,and such $$\dfrac{95}{36}>\dfrac{p}{q}>\dfrac{96}{37}$$

Find the minimum of the $q$

maybe can use $$95q>36p$$ and $$37p>96q$$ and then find this minimum of the value?

before I find a $$2.638\approx \dfrac{95}{36}>\dfrac{49}{18}\approx 2.722>\dfrac{96}{37}\approx 2.59 $$ is not such condition

idea 2: since $$\dfrac{95}{36}=\dfrac{95\cdot 37}{36\cdot 37}=\dfrac{3515}{1332}$$ $$\dfrac{96}{37}=\dfrac{96\cdot 36}{36\cdot 37}=\dfrac{3456}{1332}$$ so $$\dfrac{3515}{1332}>\dfrac{p}{q}>\dfrac{3456}{1332}$$ so $$p\in(3456,3515),q=1332$$

Answer 1

An interesting trick so solve such kind of problems is to consider the continued fraction of the LHS and the RHS. We have: $$\frac{95}{36}=[2;1,1,1,3,3],\qquad \frac{96}{37}=[2;1,1,2,7]$$ hence $$\frac{13}{5}=[2;1,1,2]$$ just lies between the LHS and the RHS, and it is the rational number with the smallest denominator lying in that interval.

Answer 2

We have $$2.64\gt a=\frac{17575}{5\cdot 36\cdot 37}=\frac{95}{36}\gt \color{red}{\frac{13}{5}}=2.6=\frac{17316}{5\cdot 36\cdot 37}\gt\frac{96}{37}=\frac{17280}{5\cdot 36\cdot 37}=b\gt 2.59.$$ Note that $$\frac{11}{4}\gt a\gt b\gt\frac{10}{4}$$

$$\frac{8}{3}\gt a\gt b\gt\frac{7}{3}$$

$$\frac{6}{2}\gt a\gt b\gt\frac{5}{2}$$ $$\frac{3}{1}\gt a\gt b\gt\frac{2}{1}$$ Hence, the minimum of $q$ is $5$.