2. Find the smallest possible value of $\left|2015 m^{5}-2014 n^{4}\right|$, given that $m, n$ are natural numbers.
Answer: 0.

Solution: Let's find $N=2014^{x} \cdot 2015^{y}$ such that $m^{5}=2014^{x-1} \cdot 2015^{y}$ and $n^{4}=2014^{x} \cdot 2015^{y-1}$. For this, $x$ and $y-1$ must be multiples of 4, and $x-1$ and $y-5$ must be multiples of 5. For example, $x=16$ and $y=5$ work. Then, if we take $m=2014^{3} \cdot 2015$ and $n=2014^{4} \cdot 2015$, we get $\left|2015 m^{5}-2014 n^{4}\right|=0$.