## Task 4 - 210514

A problem from a Leningrad Mathematical Olympiad:

An "October Child" (that is, a Young Pioneer up to 3rd grade), a Pioneer, and a Komsomol member went to a Pioneer camp. Their first names are (not necessarily in the same order) Kolya, Igor, and Sasha. From their conversations on the train, we learned:

(1) Kolya and the Komsomol member are two enthusiastic anglers.

(2) The October Child lives in Leningrad; Sasha also lives in Leningrad, but on a different street.

(3) Sasha is younger than the Komsomol member.

What is the first name of the October Child, the first name of the Pioneer, and the first name of the Komsomol member?
From (1) it follows: The Komsomol member is not named Kolya.

From (3) it follows: The Komsomol member is not named Sasha. Therefore, he has the first name Igor. Hence, the October Child is not named Igor.

From (2) it follows: The October Child is not named Sasha. Therefore, it has the first name Kolya. This leaves only the first name Sasha for the Pioneer.

Solutions of the I. Round 1981 adopted from $[5]$

### 2.22.2 II. Round 1981, Class 5