## Task 19/68

In the year 1968, someone is exactly as old as the sum of the digits of their birth year. In which year was he born?
The year of birth must obviously lie in the 20th century; if it were to fall into the 19th century, the highest possible sum of the digits would be that of the year 1899, which is 27. However, $1899+27<1968$.

Therefore, one can set up the equation

$$
1900+10z+e+(1+9+z+e)=1968
$$

where $z$ is the tens digit and $e$ is the units digit of the year of birth. This simplifies to $z=\frac{2}{11}(29-e)$.

Since $e$ must satisfy the inequality $0 \leq e \leq 9$ and $z$ must be an integer (due to their definitions as the tens and units digits, respectively), only $e=7$ is possible. This makes $z=4$.

Thus, the year of birth is 1947, with a digit sum of 21, which is exactly the age in the year 1968.