A       B
+-----+
|        |
|        |
+-----+
D       C

Precisely one stone is situated initially at each of the four vertices of the square ABCD. It is permissible to change the number of stones according to the following rule:

Any move consists of taking n stones away from any vertex and adding 2n stones to either adjacent vertex.

Prove that it is not possible to get 1989, 1988, 1990 and 1989 stones respectively at the vertices A, B, C and D after a finite number of moves.

Some things to note about allowable moves:

Only one vertex can change parity on a given turn.
The total number of stones can only increase.
The total number of stones is 7956 = 4*1989.
The total number of stones must be increased by 7956-4 = 7952.
To increase the number of stones by a total of 7952, a total of 3976 must be removed.
Removing n stones is equivalent to removing 1 stone n times.
The number of one stone moves must be exactly 3976.
A one stone move changes the difference between two adjacent verticies by 3 (or -3) and the other differences change by 1 and -2 (or -1 and 2)
The total change of differences after a one stone move is either
-3+1-2 = -4
-3-1+2 = -2
3+1-2 = 2
3-1+2 = 4
The starting differences are 0,0,0,0 which sums to 0
The goal has differences of 1,2,1,0 which sums to 4.

I keep thinking I'm close but I haven't gotten there yet.

Posted by Jer on 2010-05-05 16:20:55