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 The Central Cell (Posted on 2004-04-16)
Prove that the central cell (the number in the middle cell) of any 3x3 magic square is always one-third the magic constant (the sum of any side, either 2 major diagonals, or either center row in the magic square).
Show that in any larger square (n x n), the central cell does not need to be 1/n the magic constant.

 See The Solution Submitted by Victor Zapana Rating: 4.5000 (4 votes)

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 re: General counter-example | Comment 8 of 11 |
(In reply to General counter-example by Tristan)

I like the idea of a general counter-example, but I'd like to make one that is easier to prove.

Start off with the same 5 x 5 square.

01000
00001
00100
10000
00010

This time, to regenerate, there is no need to turn upside-down.  New columns will be added before and after the middle column.  New rows will be added before and after the middle three rows.

The next two squares following this regeneration sequence:

0100000
0010000
0000001
0001000
1000000
0000100
0000010

010000000
001000000
000100000
000000001
000010000
100000000
000001000
000000100
000000010

Proof:

It is easy to see why all rows and columns will add up to 1. The new rows and columns do not interfere with the old ones, and have a sum of 1 themselves.

For the major diagonals, the central 1 will always stay in the center, always keeping both diagonals' sums at 1.  There won't be any other 1s in the diagonals because the regeneration sequence never puts them there, nor moves them there.  The 1s on the very left and right only move further left and right.  The other 1s move diagonally, and so cannot move onto the diagonal.

For the sake of completeness, I quote my previous comment:

"If every number must be different, than add the counter-example to a magic square filled with even numbers to get your new counter-example."

 Posted by Tristan on 2004-04-16 18:32:53

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