7.1. Find two natural numbers $m$ and $n$, given that $m<n<2m$ and $mn=2013$.
Answer: $\quad m=33, n=61$.

The solution follows from the prime factorization $2013=3 \cdot 11 \cdot 61$ and a small enumeration of the divisors of 2013; there are 8 in total, and the "candidates" for the role of the number $m$ are the smaller divisors, namely the four numbers $1,3,11,33$, and among them, only $m=33$ satisfies the condition $m<n<2m$ with $n=61$.