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Successive Stones Settlement (Posted on 2010-05-03) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
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Solution Solution (Spoiler) Comment 2 of 2 |
Representing the number of stones at each vertex in modulo 3 arithmetic, the target situation (1989,1988,1990,1989) becomes (0,2,1,0).

The notation [0,2,1,0] will be used to represent (0,2,1,0) or any of its cyclical rotations (2,1,0,0), (1,0,0,2), (0,0,2,1).

Any move is equivalent to a succession of elementary moves, each involving the removal of just one stone and the addition of two.

The following table shows all possible outcomes when 9 particular representations, labelled P to X, are subjected to any elementary move.

                                    Possible outcomes
P:   [0,0,0,0]      ->         R
Q:   [0,0,1,1]     ->         P, S or W
R:   [0,0,2,2]     ->         Q, U or X
S:   [0,1,0,2]     ->         R or T
T:   [0,1,2,1]      ->         Q or U
U:   [0,2,1,2]     ->         S or W
V:   [1,1,1,1]     ->         Q
W:   [1,1,2,2]     ->         R, T or V
X:   [2,2,2,2]     ->         W

Since the starting position is represented by V, and since all the outcomes themselves also belong to the set of nine representations listed above, it follows that any succession of moves can never lead to the target [0,2,1,0] since it does not belong to this set of nine.

  Posted by Harry on 2010-05-07 22:24:49
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