Every cell in the 6x6 grid is divided into four segments, each of which contains one digit. Only one of these digits is correct; the other three are red herrings! In the three starter cells (GA, LC and LE), the rogue numbers have been shaded.
Can you shade the rogue numbers in all the other cells, so that the finished grid has the numbers 1 to 6, once only, in every row, column and coloured rectangle?
| A | | B | | C | | D | | E | | F | | |
G | 1 | 5 | 3 | 2 | 5 | 3 | 4 | 1 | 3 | 6 | 1 | 4 | |
| 2 | 3 | 1 | 4 | 1 | 2 | 6 | 5 | 4 | 1 | 5 | 2 | |
H | 4 | 5 | 2 | 3 | 4 | 3 | 5 | 1 | 2 | 3 | 6 | 5 | |
| 2 | 6 | 1 | 5 | 2 | 1 | 3 | 4 | 1 | 5 | 4 | 3 | |
I | 6 | 1 | 2 | 3 | 1 | 5 | 2 | 6 | 5 | 4 | 3 | 6 | |
| 3 | 2 | 5 | 1 | 6 | 2 | 4 | 5 | 3 | 2 | 5 | 2 | |
J | 6 | 1 | 4 | 5 | 1 | 3 | 5 | 6 | 2 | 3 | 1 | 3 | |
| 2 | 3 | 1 | 6 | 4 | 5 | 3 | 1 | 5 | 6 | 5 | 2 | |
K | 5 | 4 | 3 | 1 | 5 | 4 | 2 | 3 | 1 | 2 | 3 | 4 | |
| 6 | 1 | 4 | 5 | 2 | 6 | 5 | 4 | 3 | 4 | 5 | 6 | |
L | 1 | 2 | 5 | 4 | 3 | 4 | 6 | 2 | 6 | 1 | 2 | 1 | |
| 3 | 6 | 1 | 2 | 5 | 2 | 1 | 5 | 2 | 3 | 6 | 4 | |
The solution is given in terms of the following grid:
1...4...5...6...3...2
5...2...3...4...1...6
6...3...1...2...4...5
2...6...4...3...5...1
4...1...6...5...2...3
3...5...2...1...6...4
*** Of course, I agree that this problem is among the very best exercises dealing extensively with shapes and logic submitted by JF.
Edited on August 13, 2007, 4:10 am
Solution
