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.
(In reply to
re: Think I may have it by Quincy)
technically, i suppose the covered numbers are still upwards, even through they're covered. Of course, i'm assuming the box is lying horizontally, and that i understand what 'with the numbers upward' means...
I like Lisa's solution. It's nice and clever (besides being the only one). Way to think outside the box (haha, get it?)
re(2): Think I may have it
