2. Find any pair of natural numbers $a$ and $b$, both greater than 1, that satisfy the equation $a^{13} \cdot b^{31}=6^{2015}$.
Solution. It is enough to provide one example. Since $2015=13 \cdot 155=31 \cdot 65$, the values $a=2^{155}, b=3^{65}$ work. Indeed,

$$
\left(2^{155}\right)^{13} \cdot\left(3^{65}\right)^{31}=2^{2015} \cdot 3^{2015}=6^{2015}
$$

Comment. Besides the obvious alternative $a=3^{155}, b=2^{65}$, there can be others. For example, one can proceed as follows:

$$
\begin{gathered}
2015=5 \cdot 13 \cdot 31=2 \cdot 13 \cdot 31+3 \cdot 13 \cdot 31=62 \cdot 13+39 \cdot 31 \\
\left(6^{62}\right)^{13} \cdot\left(6^{39}\right)^{31}=6^{2015}, \quad a=6^{62}, b=6^{39} .
\end{gathered}
$$

Similarly, one can obtain the numbers $a=18^{62}=2^{62} \cdot 3^{124}, b=24^{13}=2^{39} \cdot 3^{13}$.

## Grading Criteria.

- At least one pair of values for $a$, $b$ is provided and it is shown that they satisfy the given condition - 7 points.
- A pair of numbers is provided, but nothing is justified (the jury can show that the pair works) - 5 points.
- The main idea of the solution is correct, but an arithmetic error is made (for example, it is written that $2015=13 \cdot 165$) - 2 points.