Task A-4.5. (8 points)

What is the sum of all natural numbers $n$ for which the number $\frac{2009-n}{99}$ is a natural number?
## Solution.

The number $\frac{2009-n}{99}=\frac{1980+29-n}{99}=20+\frac{29-n}{99}$ will be a natural number for

$$
n \in\{29,29+1 \cdot 99,29+2 \cdot 99, \ldots, 29+19 \cdot 99\}
$$

The sum of these numbers is $20 \cdot 29+99(1+2+\ldots+19)=580+99 \cdot \frac{19 \cdot 20}{2}=19390$.

For 4 points: correct method with a calculation error.

For 8 points: correct result and method.