(i,j) |
i → |
|
|
|
|
|
|
j ↓ |
6 |
1 |
8 |
5 |
6 |
7 |
3 |
|
7 |
6 |
6 |
5 |
9 |
9 |
2 |
|
3 |
7 |
3 |
3 |
3 |
1 |
4 |
|
7 |
8 |
3 |
4 |
2 |
1 |
7 |
|
5 |
6 |
2 |
5 |
9 |
4 |
8 |
|
4 |
2 |
9 |
4 |
3 |
6 |
5 |
|
5 |
3 |
5 |
2 |
4 |
4 |
1 |
|
6 |
9 |
6 |
8 |
6 |
8 |
9 |
|
3 |
7 |
5 |
5 |
3 |
6 |
2 |
Divide the 9x7 grid given above into seven shapes such that the numbers in each shape add up to the same total, where each of the seven shapes contain the same number of cells.(The shape of the seven parts may or may not be identical .)
Does there exist any further solutions if the seven shapes may not necessarily contain the same number of cells?
For this I copied the digits into a spreadsheet.
The sum of all digits was 315 which meant that each of the 7 shapes had to total 45.
By tallying and colour coding cells I arrived at the following arrangement:
A A A A A A A
A B B B B B A
C C C C B D D
C C C B B B D
C C E E F F D
G G E E F F D
G G E E F F D
G G G E F D D
G G E E F F D
I have not pursued an arrangement of shapes having varying cell sizes and have not needed to use an (i,j) reporting system.
Um, in Col 6 there are two 1's to which I have assigned B and D. Those appear to be the only values in my arrangement which are interchangeable .
Edited on February 9, 2012, 2:45 am
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Posted by brianjn
on 2012-02-09 02:26:49 |