Here is another game that uses an n-by-n grid drawn on paper.

This game is for 2 players.

Players take turns placing 1,2,3,... n^2 into empty squares of the grid, one number each turn.

(So, player 1 places the odd numbers into the grid, and player 2 places the evens.)

Let H represent the collection of variables h, where each h is the sum of a HORIZONTALLY-adjacent pair of integers in the grid.

Let V represent the collection of variables v, where each v is the sum of a VERTICALLY-adjacent pair of integers in the grid.

(So, H and V each consist of n*(n-1) variables.)

Player 1 then chooses a variable d which divides at least one of the h's.

Player 1 gets as a score:

d*(number of h's that are divisible by d)

Player 2 then chooses a variable d' which divides at least one of the v's.

Player 2 gets as a score:

d'*(number of v's that are divisible by d')

The player with the largest score wins.

For example, let us say two poorly-playing players have this n=4 grid:

08 09 01 10

03 11 16 07

02 14 15 05

04 12 06 13

H =

(17,10,11, 14,27,23,

16,29,20, 16,18,19)

Let us say that player 1 picks 4, then he gets 4*3 = 12 points.

(Because h's = 16, 20, and 16 are divisible by 4.)

Had he picked 29, he could have gotten 29, however.

V =

(11,5,6, 20,25,26,

17,31,21, 17,12,18)

Player 2 picks 3 and gets 3*4 = 12 points.

She should have picked 31.

Yes, you can always score at least n*(n-1) by picking d or d' = 1.

As implied in the example, a player is responsible for finding his/her own score.

Thanks,

Leroy Quet

## Sunday, June 12, 2011

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