Task B-1.4. The sum of 2011 consecutive integers is 2011. Find the largest addend.
First solution. Let $x$ be the largest addend. Then we can write

$$
\begin{aligned}
x+(x-1)+(x-2)+\ldots+(x-2010) & =2011 \\
2011 x-(1+2+3+\ldots+2010) & =2011 \\
2011 x-\frac{2010 \cdot 2011}{2} & =2011 \\
2011 x-1005 \cdot 2011 & =2011 /: 2011 \\
x-1005 & =1 \\
x & =1006 .
\end{aligned}
$$

(1 point)

The largest addend is 1006.