Andy and Bob play a game using a long straight row of squares, alternating turns. When it’s Andy’s turn, he writes an A in one of the blank squares. When Bob takes a turn, he writes a B in some blank square. (Once a letter is written in a square, neither player can use that square again.) A player wins the game when his initial is written in 4 equally-spaced squares. For example, suppose the following board is the result of several turns: (below)

     _ _ B B _ A B A _ _ _ A B _ _ A

Andy can win by writing A in the indicated square. (Four A’s with spacing 2) Bob can win by writing B in that same square. (Four B’s with spacing 3)

  • If Andy goes first, find a strategy Andy can use that guarantees that he wins. How many moves must Andy make to get 4 in a row, no matter what moves Bob makes? (Can Andy always win in just 4 moves?) Justify your answer.
  • How many squares are needed in the game board to allow Andy’s strategy to work?
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    $\begingroup$ Welcome to math.SE: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. $\endgroup$ Apr 19, 2012 at 1:20
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    – cardinal
    Apr 19, 2012 at 2:14
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    $\begingroup$ It appears that the original question about a game has been completely overwritten with a question about putting spheres in spheres. $\endgroup$ Apr 20, 2012 at 0:40
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    $\begingroup$ @Greg, I rolled it back and flagged it for moderator attention. $\endgroup$ Apr 20, 2012 at 0:43
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    $\begingroup$ @Blue Eyes, if you want to ask another question, use "ask question" on top; do not edit this question. $\endgroup$
    – sdcvvc
    Apr 20, 2012 at 1:17

1 Answer 1


Andy wins in 4 moves. Denote Andy's first move as position $0$. Without loss of generality, Bob places his move at $n$, then Andy plays at $2(n+1)$. If Bob does not play at $n+1$, $-(n+1)$ or $3(n+1)$, Andy plays at $n+1$ and wins in the next move by either playing at $-(n+1)$ or $3(n+1)$. If Bob did play one of the mentioned moves, then Andy wins by playing at $4(n+1)$ followed by either $6(n+1)$ or $-2(n+1)$.

Minimal board size looks slightly more annoying to figure out at first glance; considering that Andy's second move could happen at the opposite side of Bob's first, it looks like Bob can only prevent a spacing of $1$ and $2$, so that would mean a board size of $21$ would suffice, but I don't see a quick way to prove that without looking at several different cases, and it's entirely possible I overlooked something with that.

  • $\begingroup$ I think that with 25 board you can guarantee a win in 4 moves, and under it is not possible (but maybe you can win with more moves). If A play in x, and B in x+2, A should play his next move in x+6 or x-6, so at least 6*4+1 space are requested. $\endgroup$
    – carlop
    Apr 19, 2012 at 15:34

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