Divisor Over Sums Game

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