4.1. The number $3^{2009}$ is represented as the sum of $k$ consecutive natural numbers. What is the greatest possible value of $k$?
### 4.1. Answer: $2 \cdot 3^{1004}$

Let $3^{2009} = (a+1) + (a+2) + \cdots + (a+k) = k a + k(k+1) / 2$. Then $2 \cdot 3^{2009} = k(k + 2a + 1)$, where the first factor is less than the second. Therefore, $k < \sqrt{2 \cdot 3^{2009}} = \sqrt{6} \cdot 3^{1004}$. The largest divisor of $2 \cdot 3^{2009}$ with this property is $2 \cdot 3^{1004}$.