Task 4. Find the smallest natural $m$, for which there exists such a natural $n$ that the sets of the last 2014 digits in the decimal representation of the numbers $a=2015^{3 m+1}$ and $b=2015^{6 n+2}$ are the same, and $a<b$.
Solution. The coincidence of the last 2014 digits of the two specified powers means divisibility by \(10^{2014}\) of the difference

\[
2015^{6 n+2}-2015^{3 m+1}=2015^{3 m+1} \cdot\left(2015^{6 n-3 m+1}-1\right)
\]

since the first factor in the obtained decomposition is not divisible by 2, and the second is not divisible by 5, the divisibility of the obtained product by \(10^{2014}\) means two divisibilities at once

\[
2015^{3 m+1} \vdots 5^{2014}, \quad 2015^{6 n-3 m+1}-1 \vdots 2^{2014}
\]

The first divisibility means the inequality \(3 m+1 \geqslant 2014\), or \(m \geqslant m_{0}=671\). Let's check that the number \(m=m_{0}\) is realizable, i.e., for it, for some \(n=n_{0}\), the second divisibility also holds.

Indeed, let \(N=2^{2014}\) and note that for the odd \(K=2015\), the number

\[
K^{N}-1=\left(K^{N / 2}+1\right)\left(K^{N / 2}-1\right)=\ldots=\left(K^{N / 2}+1\right)\left(K^{N / 4}+1\right) \ldots\left(K^{1}+1\right)\left(K^{1}-1\right)
\]

is divisible by \(N\) (in the last representation of this number, all expressions in parentheses are even, and the number of parentheses is 2014). Therefore, \(\left(2015^{N}-1\right) \vdots 2^{2014}\), and the equality \(6 n_{0}-3 m_{0}+1=N\) holds for

\[
n_{0}=\left(\frac{N-1}{3}+m_{0}\right) / 2
\]

where \(n_{0} \in \mathbb{N}\), since the number

\[
N-1=2^{2014}-1=(3-1)^{2014}-1
\]

is divisible by three, and the quotient \((N-1) / 3\) is odd.

Answer: \(m=671\).