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The numbers in each of the cages are factors of the cage totals.

1. All the cages in N6 have the factor 8 -> the cells in N6 + R4C6 must be a multiple of 8 = 48 -> R4C6 = 3
1a. 3 in N6 only in R5C89 or R6C89 = {35}, 5 in R56C89, locked for N6
1b. R4C89, R5C89 and R6C89 must total either 8 or 16 so cannot contain 4,8 -> 4,8 in N6 only in R456C7, locked for C7 and N6
1c. Cage at R4C6 contains 3,4,8 and is a multiple of 8 -> must total 16 or 24 = {1348/3489}, no 2,6,7

2. All the cages in N4 have the factor 6 -> the cells in N4 + R6C4 must be a multiple of 6 = 54 (cannot be 48 because R4C6 = 3) -> R6C4 = 9
2a. 9 in N4 can only be in R5C12 = 12 = {39} (R4C12 cannot be {39} because R4C6 = 3), locked for R5 and N4
2b. 3 in N6 only in R6C89 = {35}, locked for R6 and N6

3. Similarly since all the cages in N2 + R4C4 have an even factor, R4C4 must be odd = {157}
3a. Similarly since all the cages in N8 + R6C6 have an even factor, R6C6 must be odd = {157}

4. 5-cell cage at R4C5 with factor 5 must total 25
[That also followed from the fact that R4C4, R4C6, R6C4 and R6C6 must all be odd, since if the 5-cell cage had the total 15 or 35, then R46C46 totalling 30 or 10 would need two even numbers.]
4a. 45 rule on N5 2 remaining innies R4C4 + R6C6 = 8 = {17}, locked for N5

5. All the cages in N8 + R6C6 have the factor 4 -> the cells in N8 + R6C6 must be a multiple of 4 = 52 (only possibility because R6C6 only contains 1,7) -> R6C6 = 7, R4C4 = 1

6. R6C12 must have the factor 6 = 6,12 = {24/48}, 4 locked for R6 and N4
6a. R4C12 must have the factor 6 = 12 = {57}, locked for R4 and N4
6b. R4C89 must have the factor 8 = 8 = {26}, locked for R4 and N6 -> R4C3 = 8, R4C5 = 4, R4C7 = 9
6c. Naked pair {17} in R5C89, locked for N6 -> R56C7 = [48]
6d. Naked pair {24} in R6C12, locked for R6 and N4 -> R45C3 = [61]
6e. R6C5 = 6 (hidden single in R6)

7. R789C7 must have the factor 7 = 14 (cannot total 7 or 21)
7a. 45 rule on C7 3 remaining innies R123C7 = 10
7b. R12C7 must have the factor 5 = 5 = {23}, locked for C7 and N3 -> R3C7 = 5
7c. Naked triple {167} in R789C7, locked for N9

8. R3C23 must have the factor 7 = 7 = {34} (only possibility because no 1,5,6,8 in R3C3), locked for R3 and N1

9. R23C9 must have the factor 8 = 16 = {79} (cannot be {17} which clashes with R5C9), locked for C9 and N3 -> R5C89 = [71]
9a. R1C89 must have the factor 3 = 9,12 = {18/48}, 8 locked for R1 and N3
9b. Cage at R2C8 must have the factor 4 so must have even total = 12 = {156}, 1 locked for N3, 6 locked for C8 -> R4C89 = [26]
9c. Naked pair {48} in R1C89, locked for R1

10. R78C8 must have the factor 6 = 12 = {39} (cannot be {48} which clashes with R1C8), locked for C8 and N9 -> R6C89 = [53]

11. R78C9 must have the factor 3 = 6,9 (cannot be 12 because {48} clashes with R1C9) = {24/45}, 4 locked for C9 and N9 -> R1C89 = [48]
11a. R9C8 = 8, R9C89 must have the factor 5 = 10 -> R9C9 = 2

12. R2C23 must have the factor 8 = 8 (cannot be 16 because {79} clashes with R2C9) = [17/62]
12a. Naked pair {16} in R2C28, locked for R2

13. All cells in N2 must have the factor 2, 4 in N2 only in R2C46 -> one of R12C4 and R12C6 must be [24/64] so must contain one of 2,6
13a. Cage at R3C4 must contain one of 2,6 (cannot contain both of 2,6 which clashes with R1C46), must contain one of 7,9 (cannot contain both of 7,9 which clashes with R3C9) -> must contain 8, locked for R3 and N2

14. R2C1 = 8 (hidden single in N1)
14a. 45 rule on N1 4 remaining innies R13C1 + R1C23 = 22, R13C1 has the factor 2 and is even -> R1C23 must be even and have the factor 3 = 6,12 = [15]{57}, 5 locked for R1 and N1

15. 9 in N1 only in R13C1, locked for C1 -> R5C12 = [39]
15a. R123C1 must have the factor 2 and contain 8,9 = {189/789}, no 2,6
15b. R2C23 = [62] (hidden pair in N1) -> R12C7 = [23], R23C8 = [16]
15c. R3C1 = 1 (hidden single in R3) -> R1C1 = 9
15d. Naked pair {57} in R1C23, locked for R1
15e. Naked pair {57} in R14C2, locked for C2

16. R78C1 must have the factor 5 = 10 = {46}, locked for C1 and N7 -> R6C12 = [24], R3C23 = [34], R9C2 = 1
16a. R78C1 has even total -> R7C23 must have the factor 5 and an odd total = 5,15 = [23/87]

17. R89C6 must have the factor 4 = 8,12 = [26/84]
17a. Naked pair {28} in R8C26, locked for R8
17b. R89C4 must have the factor 4 = 8,12 = {35/57}, 5 locked for C4 and N8
17c. R7C9 = 5 (hidden single in R7), R8C9 = 4, R78C1 = [46]
17d. R9C6 = 4 (hidden single in R9) -> R8C6 = 8
17e. R8C2 = 2, R7C2 = 8 -> R7C3 = 7 (step 16a), R9C1 = 5
17f. R9C7 = 6 (hidden single in R9), R78C7 = [17]

18. R1C6 = 1 (hidden single in C6) -> R1C5 = 3
18a. R89C5 must have the factor 4 = 8 = [17]

and the rest is naked singles.