(NIMO 2014) Reciprocal of Interval

Note that the reciprocal of an interval $[m,n]$ is simply the interval $[\frac1n,\frac 1m]$ . With this in mind, the condition gives $n-m=10^{10}\left(\dfrac1m-\dfrac1n\right)=10^{10}\left(\dfrac{n-m}{mn}\right)\quad\implies\quad mn=10^{10}.$ It suffices to compute the number of solutions to this Diophantine equation with $m<n$ . First remove the $m<n$ condition. Remark that if $m$ is a divisor of $10^{10}$ , then there is a unique integer $n$ such that $mn=10^{10}$ . Thus, the number of solutions to the equation is equal to the number of divisors of $10^{10}=2^{10}5^{10}$ , which is $(10+1)^2=121$ . Bringing the condition back in, we see that $m=n=10^5$ is invalid, and furthermore for every distinct $x$ and $y$ which satisfies $xy=10^{10}$ , both $(x,y)$ and $(y,x)$ are valid pairs of integers, but only one of them satisfies the $m<n$ condition. Putting everything together, we see that the final answer is $\frac{121-1}2=\boxed{60}$ .