2. Let $x_{0}, x_{1}, x_{2}, \ldots, x_{2002}$ be consecutive integers, for which

$$
-x_{0}+x_{1}-x_{2}+\ldots-x_{2000}+x_{2001}-x_{2002}=2003
$$

Calculate the number $x_{2002}$.
Solution. We rewrite equation (1) in the following form:

$$
\left(-x_{0}+x_{1}\right)+\left(-x_{2}+x_{3}\right)+\ldots+\left(-x_{2000}+x_{2001}\right)-x_{2002}=2003
$$

Since $x_{i}, i=0,1, \ldots 2002$ are consecutive integers, it follows that $x_{i+1}-x_{i}=1$, from which we obtain

$$
1001-x_{2002}=2003
$$

i.e.

$$
x_{2002}=-1002
$$