## Task 1 - 320821

Mr. Schulz, who was born in this century, notices that on his birthday in 1992, he reaches an age (in years) that is equal to four times the sum of the digits of the year of his birth.

Determine whether there is exactly one year that could be his birth year that would make Mr. Schulz's observation true! If so, name this year!
If $a$ is the units digit and $b$ is the tens digit of the year in this century, then $a$ and $b$ are natural numbers with $0 \leq a \leq 9$ and $0 \leq b \leq 9$, the year is $1900 + 10b + a$, and its digit sum is $1 + 9 + b + a$.

If such a year is Mr. Schulz's birth year, then in 1992 he reaches an age of $1992 - (1900 + 10b + a)$ years.

If Mr. Schulz's statement is true, then

$$
1992 - (1900 + 10b + a) = 4 \cdot (1 + 9 + b + a)
$$

This leads to $92 - 10b - a = 40 + 4b + 4a$ and $52 - 14b = 5a$.

If $b \geq 4$, then $5a \leq 52 - 14 \cdot 4 < 0$, which contradicts $a \geq 0$. If $b = 0$ or $b = 1$ or $b = 2$, then $5a = 52$ or $5a = 38$ or $5a = 24$, none of which can be satisfied by natural numbers $a$.

Therefore, only $b = 3$ is possible, which leads to $5a = 10$, so $a = 2$.

Thus, the statement can only be true with 1932 as the birth year.

This is indeed the case; the digit sum of 1932 is $1 + 9 + 3 + 2 = 15$, and the age reached in 1992 is $1992 - 1932 = 60 = 4 \cdot 15$. Therefore, there is exactly one year for which the statement is true; it is the year 1932.