\section*{Problem 4 - 061014}

On New Year's Day in 1953, \(A\) and \(B\) met during a train ride. In the course of their conversation, the topic of their ages came up.

\(A\) said: "If you calculate the sum of the digits of my (four-digit) birth year, you will get my age." After a brief moment of thought, \(B\) then congratulated \(A\) on his birthday.

a) How did \(B\) know, without receiving any further information, the birth date?

b) When was \(A\) born?
}

\(A\) can be at most 27 years old; for the largest cross sum possible under the given conditions is \(1+8+9+9=27\). He was thus born after 1924. Let his birth year be \(1900+10 a+b\) with \(a, b\) integers and \(2 \leq a \leq 5 ; 0 \leq b \leq 9\). His age on January 1, 1953, is therefore (according to the conditions) \(1+9+a+b\) years.

Thus, if he was born on January 1 (Case 1):

\[
\begin{aligned}
1+9+a+b & =1953-(1900+10 a+b) \\
43 & =11 a+2 b
\end{aligned}
\]

This equation, considering the conditions for \(a\) and \(b\), is only satisfied by \(a=3\) and \(b=5\). Therefore, A was born on January 1, 1935, and is 18 years old.

However, he could also have been born on another day (Case 2):

\[
\begin{aligned}
1+9+a+b & =1952-(1900+10 a+b) \\
42 & =11 a+2 b
\end{aligned}
\]

This equation is not satisfied by any pair \((a, b)\) under the conditions of the problem. The solution given for Case 1 is thus the only one.

Solutions of the I. Round 1966 adopted from [5]

\subsection*{7.8.2 II. Round 1966, Class 10}