Let Peter and Pavel be born in the years $\overline{18 x y}$ and $\overline{19 u v}$, respectively. At the time of their meeting, Peter and Pavel were $1 + 8 + x + y$ and $1 + 9 + u + v$ years old, respectively. We can determine the year of their meeting in two ways. Since Peter's age at that time was equal to the sum of the digits of his birth year, the meeting took place in the year $1800 + 10 x + y + 9 + x + y$. On the other hand, Pavel's age was also equal to the sum of the digits of his birth year, so the meeting took place in the year $1900 + 10 u + v + 10 + u + v$. Therefore, $1800 + 10 x + y + 9 + x + y = 1900 + 10 u + v + 10 + u + v$.

After simplifications, the equation transforms to $11(x - u) + 2(y - v) = 101$. We can rewrite it as $11(x - u) + 2(y - v - 1) = 99$.

From this, it is clear that $y - v - 1$ is divisible by 11. Since $-9 \leq y - v \leq 9$, then $y - v = 1$. Consequently, $x - u = 9$. Pavel is older than Peter by

$1900 + 10 u + v - 1800 - 10 x - y = 100 - 10(x - u) - (y - v) = 100 - 90 - 1 = 9$ years.

We need to consider two more cases: Peter could have been born in 1900 (which also belongs to the 19th century), or Pavel in 2000. In the first case, the meeting would have taken place in 1910, meaning Pavel was born no earlier than 1901 and no later than 1910, and he could not have been less than 11 years old at the time of the meeting. This is a contradiction. In the second case, the meeting would have taken place in 2002, and Peter would have been at least 102 years old at that time, which is also impossible, as the sum of the digits of any integer from 1801 to 1900 is no more than 27.

## Answer

9 years.
Khachaturyan A.V.

Petr was born in the 19th century, and his brother Pavel - in the 20th century. Once, the brothers met to celebrate their shared birthday. Petr said: "My age is equal to the sum of the digits of the year of my birth." - "Mine too," replied Pavel. How much younger is Pavel than Petr?