In the example below, three rectangular pieces of dimensions 2x1, with the numbers (0,5), (2,3) and (2,3) are put into the 3x3 box so that all the 6 sums (3 rows and 3 columns) are the same (=5).
+---------------------+
| +=====++=====+|
| | 2 || 3 ||
|+=====+|-----||-----||
|| 0 || 3 || 2 ||
||-----|+=====++=====+|
|| 5 | |
|+=====+ |
+---------------------+I put these six similar pieces - (0,2), (0,6), (1,1), (1,5), (2,4) and (2,4) - with the numbers upwards, in a 4x4 box and showed it to my next door neighbour. He noticed that all 8 sums (4 rows and 4 columns) added up to the same number. How did I do it?
+=====+=====+ +=====+=====+ +=====+=====+
| 0 | 2 | | 0 | 6 | | 1 | 1 |
+=====+=====+ +=====+=====+ +=====+=====+
+=====+=====+ +=====+=====+ +=====+=====+
| 1 | 5 | | 2 | 4 | | 2 | 4 |
+=====+=====+ +=====+=====+ +=====+=====+Note: the "6" shown is still a "6" even when you put that piece upside down.
All the numbers on the tiles will be counted twice (once in its row, and once in its column)...
And the numbers add up to: 0+2+0+6+1+1+1+5+2+4+2+4 = 28. So, twice that is 56. And we have 4 rows and 4 columns (8 sums) which all must be equal. So, it seems that they must all be 56/8 = 7.
Seven is odd. And the only tiles that have odd numbers on them are (1,1) and (1,5). We need an odd number in each row and column to make an odd sum for the respective row and column.
Unfortunately, I don't think we can arrange those two tiles to satisfy that condition.
So, as it is currently worded, the problem seems impossible.
Are we missing a tile or two that adds up to another 4, or perhaps different numbers on the 6 tiles?
Maybe I'm misunderstanding the problem. As always, I reserve the right to be wrong.
- SK
first thoughts...
