Here is a game for 2 players.

The game is played on a "triangle"-shaped grid. 1 square on the top

row; 2 squares on the 2nd row; 3 squares on the 3rd row; etc; n

squares on the nth and bottom row.

(I suggest an n of about 12 for beginners.)

So, we have this:

(Each @ represents a square.)

@

@ @

@ @ @

@ @ @ @

@ @ @ @ @

etc

The players take turns filling in squares of the triangle, one empty

square filled in per move.

Each player fills in floor(n^2 /8) squares. (So, 2*floor(n^2 /8)

squares are filled in by both players together.)

Player 1 gets a point for every COLUMN of the triangle where the

number of filled in squares (filled in by either player) is coprime to

the number of total squares in the column.

Player 2 gets a point for every ROW of the triangle where the number

of filled in squares is coprime to the number of total squares in the

row.

(And for the purposes of this game, 0 is considered to be coprime only

to 1.)

For example, let's say we have a triangle filled like this at the

game's end:

(n= 6. @ = empty square. X = filled square.)

X

@ X

X X @

@ @ X @

X X @ @ @

@ @ X @ @ @

Player 1: column 1 (from left), 3 is not coprime to 6 (no point);

column 2, 3 is coprime to 5 (1 point); column 3, 2 is not coprime 4

(no point); column 4, 0 is not coprime to 3 (no point); column 5, 0 is

not coprime to 2 (no point); column 6, 0 is coprime to 1 (1 point).

Player 2: row 1 (from top), 1 is coprime to 1 (1 point); row 2, 1 is

coprime to 2 (1 point); row 3, 2 is coprime to 3 (1 point); row 4, 1

is coprime to 4 (1 point); row 5, 2 is coprime to 5 (1 point); row 6,

1 is coprime to 6 (1 point).

So, player 1 gets 2 points, while player 2 gets 6 points. Player 2

wins.

Is there a strategy that guarantees a win for one player?

Thanks,

Leroy Quet

## Monday, September 22, 2008

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