It is known that $A$ is the largest of the numbers that are the product of several natural numbers whose sum is 2011.

What is the highest power of three that divides the number $A$?
1) Note that among the factors included in $A$, there are no ones. Indeed, in the representation of the number 2011 as a sum, two addends 1 and $a$ can be replaced by one addend $1+a$, and the product will increase: $1+a > 1 \cdot a$.
2) We will prove that among the factors included in $A$, there are no numbers greater than 4. Indeed, if $A \geq 5$, then $2a - 9 > 0$, which means $a < 3(a-3)$, and by replacing the addend $a$ with two addends $a-3$ and 3, we can keep the sum and increase the product.
3) If we replace 4 with two twos, neither the sum nor the product will change. Therefore, we can assume that there are no fours among the factors included in $A$. Thus, in the product $A$, only the factors 2 and 3 appear.
4) There can be no more than two twos, since $2+2+2=3+3$, and $2^3 < 3^2$.
5) The numbers 2011 and $2009=2011-2$ are not divisible by 3, while the number $2007=2011-2-2$ is divisible by 3. Therefore, in the product $A$, there are exactly two twos. Thus, $A=2^2 \cdot 3^{669}$.

## Answer

$3^{669}$.