$a,b$ are positive integers. Let $\frac{a}{b}$ be the fraction with the smallest possible denominator $b$ such that $\frac{386}{2019}$ < $\frac{a}{b}$ < $\frac{35}{183}$. Determine the value of $a+b$.
I have tried simplifying the inequality, but I am stuck. However, I do know that as $b$ has to be smallest, so does $a$.
Any idea how I should do this question? Thanks for any help.
Maybe the following will help.
We have $$386b+1\leq2019a$$ and $$35b\geq183a+1.$$ We can solve the equation $35b=183a+1,$ which gives $$(a,b)=(13+35k,68+183k),$$ where $k\geq0$ is an integer, which gives a fraction $\frac{13}{68}.$
Easy to see that $\frac{13}{68}$ is not valid.
Now, we can take $k=1$, $k=2$,...
Also, we can solve the equation $386b+1=2019a,$ which gives $$(a,b)=(373+386k,1951+2019k),$$ where $k\geq0$ is integer.
Easy to see that $\frac{373}{1951}$ is valid.
I got that in the first case $k=1$ is valid, which gives $\frac{48}{251}.$
The continued fraction of $386/2019$ is $[0; 5, 4, 2, 1, 29]$.
The continued fraction of $35/183$ is $[0; 5, 4, 2, 1, 2]$.
So the simplest fraction that lies strictly between these numbers has continued fraction $$[0; 5, 4, 2, 1, 3]=\dfrac{48}{251}$$
