[Anonymous314]

Let $m,n$ be positive integers such that $\displaystyle{\frac{108}{997}<\frac{m}{n}<\frac{110}{999}}$.
Find the minimum value of $m+n$

Solution without approximation.
$$\frac{108}{997}<\frac{m}{n}<\frac{110}{999}\leftrightarrow \frac{999}{110}<\frac{n}{m}<\frac{997}{108}\leftrightarrow 9+\frac{9}{110}<\frac{n}{m}<9+\frac{25}{108}$$
We have $$9<\frac{n}{m}<10\longleftrightarrow 9m<n<10m.$$
Thus $n$ can be expressed in $9m+x$ for $x \in \mathbb{N}$ and $0<x<m$.
We have $$\frac{9}{110}<\frac{x}{m}<\frac{25}{108}$$
$$\frac{110}{9}>\frac{m}{x}>\frac{108}{25}$$
$$\frac{110x}{9}>m>\frac{108x}{25}$$
but we want to find the minimum value of $m+n$, that is we want to find the minimum value of $m,n$ that satisties the equation.
Thus $x=1$ and $\displaystyle{\frac{110}{9}>m>\frac{108}{25}}$, and we get $\min(m)=5$.
From $n=9m+x$, we have $n=9\cdot 5 +1=46$.

$$\therefore \frac{m}{n}=\frac{5}{46}$$

Thus the minimum value of $m+n$ that satisfies the equation: $\displaystyle{\frac{108}{997}<\frac{m}{n}<\frac{110}{999}}$ is $m+n=5+46=51$
as desired ##.