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Problem 1664

National Olympiad, first round

Combinatorics Difficulty 6.3 Prove it Russian Mathematical Olympiad · Russia

7 cards with numbers $0$ , $1$ , $2$ , $3$ , $4$ , $5$ , $6$ are given. Peter and Basil make moves in turn taking one card by each move; Peter makes the first move. The player who can construct of his cards a decimal number divisible by $17$ earlier than his opponent is declared as a Winner. Determine which of two players has a winning strategy. (I. Rubanov)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let us denote the players as $A$ (the first player) and $B$ (his opponent).

We present a strategy that allows $A$ to guarantee a win. Let him take the digit $3$ on his first move; then $B$ is forced to take $4$ (otherwise $A$ will take it on his second move and win by forming the number $34$ ). Note that $B$ cannot win on his second move, since the only two-digit number containing $4$ in its digits and divisible by $17$ is $34$ .

Next, $A$ takes $1$ , then $B$ must take $5$ (indeed, otherwise he will not win with this move, and on the next move $A$ will take $5$ and form $51$ ). Then, on the next move, $A$ takes $6$ and wins by forming the number $136$ .

Remark. There exist other winning strategies for $A$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.