(Quantum magazine)
All irreducible fractions whose denominators do not exceed 99 are written in ascending order from left to right. What are the fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ on each side of $\dfrac{5}{8}?$
I have no idea to solve this problem. Can someone give me a hint? Thanks for antetion
First of all, it should be clear that once we solve the problem for one side, the second side will be done similarly.
So let $$\frac{a}{b} < \frac{5}{8}\tag{1}$$ with $b \leq 99$. Now, let $a$ be fixed, then from $(1)$ we get $$\frac{8a}{5}<b\tag{2}.$$ In order to maximize $a/b$, we want to minimize $b$ for fixed $a$, so we want to choose next integer after $8a/5$. To see what integer this will be, we can inspect $5$ cases based on remainder of $a$ divided by $5$.
For example, if $a \equiv 0 \bmod {5}$, then $(2)$ gives us $b \geq \frac{8a}{5}+1$, and so $b$ is minimized for $b=\frac{8a}{5}+1$. But then considering $a$ was chosen arbitrary, we will choose maximal $a$ such that $b=\frac{8a}{5}+1\leq 99$ and $a \equiv 0 \bmod {5}$. The inequality simplifies to $a \leq 61$ (since $a$ is an integer), and so maximal $a$ is with $a=60$, and thus $b=97$.
Doing this process for all $a\equiv 0,1,2,3,4 \bmod 5$, we will obtain $\frac{a}{b}=\frac{60}{97}$,$\frac{61}{98}$,$\frac{57}{92}$,$\frac{58}{93}$,$\frac{59}{95}$, respectively. Directly comparing the five possibilities, we can see that maximum is at $$\frac{a}{b}=\frac{58}{93}.$$