\section*{Problem 3 - 291013}

Two players have invented the following game: On a game board, 14 fields are arranged in a circle, one of these fields is designated as the starting field \(A\). Each player has a game piece and places it on field \(A\).

Then each player rolls a die. If both players roll different numbers, the player with the higher number moves their piece 4 steps clockwise, the other player moves their piece 2 steps. If they roll the same number, each player moves their piece 3 steps forward.

This rolling and moving of both pieces is considered one turn. Due to the circular arrangement of the fields, it is possible for a piece to reach or pass field \(A\) (and thus start a new lap) during its movement.

The game is over as soon as after performing a turn, the piece of at least one player lands exactly on field \(A\). This player wins if the other player's piece is not on \(A\). However, if both pieces are on \(A\), the game ends in a draw.

What is the smallest possible number of turns in which a game can end in a draw? Justify your answer!
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I. If a game ends in a draw after \(n\) "moves", the sum \(s\) of the total number of steps taken by both stones is 6n. Furthermore, one player's stone has completed a number \(a\) of full laps, and the other player's stone has completed a number \(b\) of full laps, so it follows that

\[
s=14 a+14 b=14(a+b)
\]

Thus, \(s\) is a common multiple of 6 and 14 and therefore a multiple of the least common multiple 42 of the numbers 6 and 14; i.e., \(s=42 g\) with an integer \(g\).

Hence, \(6 n=42 g, n=7 g\), i.e., \(n\) is divisible by 7. Therefore, a draw cannot occur in fewer than 7 "moves".

II. A draw can occur in 7 "moves", for example, if one player always takes 2 steps in these 7 "moves" and the other player always takes 4 steps (since they always roll the higher number).

With I. and II., it is proven: The smallest possible number of "moves" in which a draw can occur is 7.