1. We took 2015 consecutive natural numbers, divisible by 2015. Can their sum be the 2015th power of some natural number?

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# Solution.

Let the first of the given numbers be denoted as \( a \). Then the second number is \( a + 2015 \), the third number is \( a + 2015 \cdot 2 \), and so on. The last number is \( a + 2015 \cdot 2014 \). These numbers form an arithmetic progression with a common difference of 2015. Their sum \( S \) is

\[
\frac{2a + 2015 \cdot 2014}{2}
\]

Let's take, for example, \( a = 2015^{2014} - 1007 \cdot 2015 \). Then \( S = 2015^{2015} \).

Answer: it can.

## Recommendations for checking.

Answer without specifying the numbers - 0 points.

Any example showing that such numbers exist - 7 points.