1. This puzzle is quite hard due to the number
of large cages in it. Start by looking at what cannot be in the large
cages. For example, the 42-cage in the upper right cannot contain 1
or 2. And the 39-cage in the upper-right cannot contain a 1, 2, or
3. (since it is a 6-cage summing to 39). Thus, the red cells must
contain 1 and 2.
2. And, since the 42-cage cannot contain 1 or
2 and the 26-cage with the red cells already has 1 and 2 in it, the
blue cell must be 1 or 2.
3. The rule of 45 indicates the blue cells must be 27 or 28.
If they sum to 28, then they are:
- 4789 or
- 5689
and, then red cells contain 1 and 3 and something else.
If they sum to 27, then they are:
- 4689 or
- 5679
and, then the red cells contain 2 and 3 and something
else.
4. The 26-cage has 4 unknown cells. They sum to 23 with a 1 or 2:
- 3479
- 3569
- 3578
- 4568
If the 8-cage (in red) is 134, then the blue cell must contain
a 3 or 4. If the 8-cage is 125, then the the blue cell must
contain a 5 or 3479.
5. The blue cells must sum to 14 (rule of 45
combined with knowing the red cells will both be 1 or both be
2.
6. The blue cell at C5R6 cannot contain a 9.
If it did then, the blue cell at C9R5 would have to a 9. and the only
possible total for the red cells of the 28-cage would be 26 (the only
way not to include a 9). But in that case there is no cell in row
six which could contain a 3.
7. By the rule of 45, the blue cells must be
23 more than the reed cell. Thus, the red cell must be a 1.
8. From this we can now set the red
cells. And, we can remove 1 from the blue cells.
9. Since the red cells must sum to 27 (and they cannot contain a 3) we know the top three cells must contain a 6.
Thus, the blue cells, which also sum to 27, cannot contain a 6.
They must then be 3789.
10. Due to the 3 in the 38-cage, we know that
the red cell cannot have a 3.
11. The red cells sum to 14. (See step 5). It
cannot contain 1, 3, or 6. This leaves: 257 or 248. However, 248
would be illegal, since the blue cells must contain an 8. Thus, the
red cells must be 2+5+7. Furthermore, the 7 must be in column 5, due
to the need for a 7 in the blue cells.
12. Some clean up.
13. Since the 42-cage cannot contain a 2 and
the 28-cage (both in the upper-left nonet) already has a 2, we know
the 8-cage must be 125.
14. The red cells sum to 21 and they cannot
contain 1, 2, or 5. Thus, they must be either: 489 or 678. In other
words, they must contain an 8. This means the blue region must
contain an 8. And the 26-cage to the right of the 42-cage cannot
contain an 8. Since it already has a 1, 2, and 7 the only legal sum
is 123479.
15. We can now eliminate lots of possibilities
in the top left and top middle nonet.
16. The red cells force the blue cell to be a
1.
17. The use of 2357 in the sixth row limits
the possible sum of the red region to 4689.
18. The rule of 45 makes the red cell equal to
the blue cell.
19. The blue cells force the cells in the
third column of the lower left nonet to contain 25. Since the bottom
cell doesn't contain either the red cells must be 25.
20. The rule of neccesity forces the red cell
to be a 1.
21. The only place left for a 1 in the bottom
left nonet is the red cell.
22. The red cells in the middle left nonet
must contain a 4 and 6. Thus, the blue cells must also contain a 4
and 6. This forces the blue cells, which must sum to 19 to be
469.
23. Some clean up. The red cell cannot be a
9. We know then (from step 18) that the blue cell cannot be a
9.
24. The rule of necessity makes the red cell a
4. Likewise, the blue cells must contain a 6.
25. The only place for a 1 in the middle right
nonet is the red cell.
26. The bottom midle nonet must have a 1 in
the red cells. Which means the bottom right nonet must have a 1 in
the blue cell.
27. The red cells in the middle nonet must sum to 17. The only valid sums are:
- 359
- 368
- 458
Thus the blue cell must be 8 or 9. And the a 2 must be in row 5.
28. The red cells in the middle nonet must sum to 17. The only valid sums are:
- 359
- 368
- 458
Thus the blue cell must be 8 or 9. And the a 2 must be in row
5.
29. Using the information we have we can fill
in the possibilities for the blue cells.
30. The 28-cage in red cannot contain 1 or 7.
Thus, it must be 234568. From this we can see that the red cell
cannot be an 8.
31. The red cells eliminate the possibility of
a 9 in the blue cell.
32. The red cells sum to 24. The possible sums are:
- 2589
- 2679 Not possible
- 3489 Not possible
- 3579 Not possible
- 3678
- 4569 Not possible
- 4578
The remaining possible sums are:
- 2589
- 3678
- 4578
Looking at these we see that the only place for a 7 in the bottom nonet is in the blue cells.
33. The possible sums also determine that the
red cell cannot contain either 3 or 5.
34. To make the 29-cage on the left the red cells must be 46 and either 3 or 8. The blue cells must be 7 and either 3 or 8.
35. The red cells must sum to 11 (45-(24+10)),
so C1R5 must be 4 or 6.
36. The red cells eliminate an 8 from the blue
cell.
37. The rule of 45 indicates that the red
cells have to be 9 more than the blue cell. Using this we can see
that there is no possible way for a 9 to be in C1R9.
38. Some basic elimination can now be
applied.
39. The red cells must sum to 14.
40. Basic elimination
41. Rule of necessity.
42. The blue cells sum to 21, so the 2 is
illegal. And, then there is only one way to finish this
cage.
43. Basic Eliminationbbb77
44. The 28-cage in red cannot contain a 9, so
the blue region must contain a 9. Since the 11-cage to the right
cannot contain a 9, we know that a 9 must be in one of the cells in
the red cells of the bottom right nonet. If we look at the possible
sums for 26-cage on the right we see it must include 469 in the red
cells.
45. This means the 11-cage must be
137
46. This leaves 258 for the red region and 379
for the blue region.
47. Basic Elimination.bbb
48. Rule of necessity and basic
elimination
49. If you have followed along to this point,
the rest of the puzzle should fall into place.
