After reviewing Charlie’s set of 9 generic solution to my
Magic matrix
I have found a common feature to all solutions:
The central cell of the matrix is always a multiple of 3.
Show, why it is a must.
From memory, the earlier problem asked for the number of arrangements of digits 1-9 in 3x3 grid such that each row, column, and diagonal is divisible by 9.
Add up the row, column, and both diagonals passing through the central square. That total will be divisible by 9 and counts every digit once and the central digit an additional three times, or (sum of all) + (3 * central cell) is divisible by 9. Since (sum of all)=45 which is divisible by 9, (3 * central cell) will be divisible by 9 as well, requiring central cell=3,6, or 9.
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Posted by xdog
on 2015-11-04 12:24:34 |
No Subject