At this point, the puzzle is jammed …
A | B | C | D | E | F | G | H | I | |
1 | 7 | 6 | 1 | 8 | 9 | ||||
2 | 6 | 4 | 7 | 2 | |||||
3 | 1 | 9 | 6 | ||||||
4 | 6 | 8 | 7 | 1 | 4 | ||||
5 | 1 | 4 | 6 | 2 | |||||
6 | 7 | 4 | 9 | 5 | 3 | 6 | 1 | ||
7 | 6 | 9 | 3 | 4 | |||||
8 | 4 | 1 | 6 | ||||||
9 | 5 | 7 | 6 |
13 I5 = 7 C
… And the puzzle is jammed again …
14 I9 = 8 C
… And can you believe it? After being jammed twice, it is now locked!
A | B | C | D | E | F | G | H | I | |
1 | 7 | 6 | 1 | 8 | 9 | ||||
2 | 6 | 4 | 7 | 2 | |||||
3 | 1 | 9 | 6 | ||||||
4 | 6 | 8 | 7 | 1 | 4 | ||||
5 | 1 | 4 | 6 | 2 | 7 | ||||
6 | 7 | 4 | 9 | 5 | 3 | 6 | 1 | ||
7 | 6 | 9 | 3 | 4 | |||||
8 | 4 | 1 | 6 | ||||||
9 | 5 | 7 | 6 | 8 |
There is a full house in column A, because cells A6 and A7 may only house the digits 2 and 8. Therefore, the digits 2 and 8 can be eliminated from the other cells in the column (A2, A5 and A9). Consequently, 3 is the only possible digit in A9.
15 A67 : 28 N
16 A9 / 2 C
17 A2 / 8 C
18 A5 / 8 C
19 A9 = 3 N
… And yet locked again! (Click here for continuation - Part 3)
A | B | C | D | E | F | G | H | I | |
1 | 7 | 6 | 1 | 8 | 9 | ||||
2 | 6 | 4 | 7 | 2 | |||||
3 | 1 | 9 | 6 | ||||||
4 | 6 | 8 | 7 | 1 | 4 | ||||
5 | 1 | 4 | 6 | 2 | 7 | ||||
6 | 28 | 7 | 4 | 9 | 5 | 3 | 6 | 1 | |
7 | 28 | 6 | 9 | 3 | 4 | ||||
8 | 4 | 1 | 6 | ||||||
9 | 3 | 5 | 7 | 6 | 8 |
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