Minimize $m+n$ given $\frac{2016}{2017}<\frac mn<\frac{2017}{2018}$ Given:
$$\dfrac{2016}{2017}<\dfrac mn<\dfrac{2017}{2018}$$
Find the smallest value possible of the sum of the denominator and the numerator, i.e. $m+n$.
I don't know how to spot the very peculiar fraction with the minimum values of $m$ and $n$ in the domain $\left[\dfrac{2016}{2017},\dfrac{2017}{2018}\right]$.
Edit: thank you everyone for the answers! 
how can one prove that using the median operator the result that is strictly between the two fractions is the one with the minimized value possible for m+n.
I've done some work, I realized that I need to count the number of digits in a decimal number c, given x < c < y , let's say N, c= c*10^(N-1)/10^(N-1)
m+n= c*10^(N-1)/(gcd(c*10^(N-1),10^(N-1))) + 10^(N-1)/(gcd(c*10^(N-1),10^(N-1)))
m+n= (c+1)(  10^(N-1) / (gcd(c*10^(N-1),10^(N-1)) )
c is a variable that changes on the domain (x,y) and N is dependent on c, so N is also a variable, and then (m+n) is the last variable that is dependent on N and the gcd, which means on both N and c, to write N in terms of c for natural numbers, it's pretty easy and straight-forward: N= ceiling(log(c)) or N= floor(log(c))+1, now, since x and y are at least consecutive numbers, the variable c won't be a natural number and its length isn't easily given.
1) how can you be determined of the number of digits in any decimal number including the fractional part?
2) is it possible to have a function with two variables f(x,y), from which we can obtain the minimum value of m+n? 
thanks a ton! I look forward for your answers.

$$\frac{1}{2016}>\frac{m}{n}-1>\frac{1}{2017}$$
$$\frac{1}{2016}>\frac{m-n}{n}>\frac{1}{2017}$$
$$2016<\frac{n}{m-n}<2017$$
So $n\ge 2\cdot2016+1=4033$ and $m-n\ge 2$. Hence $m\ge 4035$
The smallest value is $8068$.
