2. Vasya arranges natural numbers from 1 to 10 in the product \(a^{b} b^{c} c^{d} d^{e} e^{f} f^{g} g^{k} k^{l} l^{m} m^{a}\) (identical letters denote identical digits, different letters denote different digits). What is the highest power of two that can divide this product?
Answer: $2^{69}$.

Solution. It is clear that in the highest possible power (10) there should be the digit $8=2^{3}$. In the power of 9, there should only be $4=2^{2}$, and the other even digits should be in the powers of 8, 7, and 6. Then the product will be divisible by a power of two equal to $3 \cdot 10+2 \cdot 9+8+7+6=69$. At the same time, such a variant is quite possible, for example, $2^{8} \cdot 8^{10} \cdot 10^{6} \cdot 6^{7} \cdot 7^{4} \cdot 4^{9} \cdot 9^{1} \cdot 1^{3} \cdot 3^{5} \cdot 5^{2}$.

Comment. It is determined that in the highest possible power, the digit 8 should be placed - 1 point. The exponents for 8, 4, 2, 6, 10 are determined - 4 points. The correct answer about the power of two is found and justified - 5 points. An example of a number is provided - another 2 points.