3. Find such an integer $\mathrm{n}$ that for any integer $\mathrm{k} \neq 2013$, the number $n-k^{2014}$ is divisible by $2013-k$.

Answer: $2013^{2014}$.
# Solution

Let $t=2013-k$, then $\frac{n-k^{2014}}{2013-k}=\frac{n-(2013-t)^{2014}}{t}$. If we raise the number $2013-t$ to the power of 2014, we get a sum of terms, each of which contains a factor $t^{m}$ (for some $m=1, \ldots, 2014$) except for $2013^{2014}$, then we get $\frac{n-(2013-t)^{2014}}{t}=\frac{n-2013^{2014}-t S}{t}=\frac{n-2013^{2014}}{t}-S$, where S is an integer, meaning $n-2013^{2014}: t$ for any integer t. This is only possible when $n-2013^{2014}=0 \Rightarrow n=2013^{2014}$.

## Recommendations for checking

1) It is proven that the number $\mathrm{n}=2013^{2014}$ satisfies the condition of the problem, but it is not proven that there are no other numbers that satisfy the condition - 2 points.