2. A person will turn as many years old in 2008 as the sum of the digits of their birth year. In which year was this person born?
2. Let the birth year of the person be denoted by $\overline{a b c d}$. It is easily concluded that then

2 POINTS

1) $a=1$ and $b=9$ or
2) $a=2$ and $b=c=0$.

In the first case,

$$
(1900+10 c+d)+(a+b+c+d)=2008
$$

i.e.,

$$
11 c+2 d=98
$$

from which it follows that $c=8$ and $d=5$. Therefore, one solution is 1985, i.e., the person was born in 1985.

In the second case,

$$
(2000+d)+(2+d)=2008
$$

from which it follows that $2 d=6$, i.e., $d=3$. Thus, the second solution is 2003, i.e., the person was born in 2003.

TOTAL

4 POINTS 10 POINTS

$$
(1+2007)+(3+2005)+\cdots+(1003+1005)=2008 \cdot 502
$$

4 POINTS

The denominator of the left side has the value:

$$
2 \cdot(1+2+3+\cdots+1003+1004)=2 \cdot \frac{1004}{2} \cdot 1005=1004 \cdot 1005
$$

4 POINTS

The equation becomes:

$$
\begin{aligned}
\frac{2008 \cdot 502}{1004 \cdot 1005} & =\frac{x}{2010} \\
\frac{1004}{1005} & =\frac{x}{2010} \\
\frac{2008}{2010} & =\frac{x}{2010} \\
x & =2008
\end{aligned}
$$