Sudoku Solution No. 109

Hard Score 72 / 100
Digits appear one by one.

Step-by-step solution

Steps 64 Estimated time 104 – 156 min Hardest technique X-Wing Clues 23
Given digit (clue) Solved digit Placement (✓ RlCc=d) Elimination (✗ RlCcd)
1 Hidden Single ✓ R1C7=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R1C7 = 1.
2 Hidden Single ✓ R3C7=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R3C7 = 2.
3 Hidden Single ✓ R3C9=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R3C9 = 9.
4 Hidden Single ✓ R5C1=9
Scan the empty cells in this column where 9 could go. Across the whole column, 9 has only one possible spot left. So R5C1 = 9.
5 Hidden Single ✓ R7C6=6
Scan the empty cells in this row where 6 could go. Across the whole row, 6 has only one possible spot left. So R7C6 = 6.
6 Hidden Single ✓ R7C8=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R7C8 = 9.
7 Hidden Single ✓ R8C1=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R8C1 = 3.
8 Hidden Single ✓ R8C3=2
Scan the empty cells in this column where 2 could go. Across the whole column, 2 has only one possible spot left. So R8C3 = 2.
9 Naked Single ✓ R8C4=9
Look at cell R8C4. Every other digit already appears in its row, column or box. Only one option remains: R8C4 = 9.
10 Hidden Single ✓ R1C6=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R1C6 = 3.
11 Hidden Single ✓ R2C3=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R2C3 = 4.
12 Hidden Single ✓ R2C4=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R2C4 = 7.
13 Hidden Single ✓ R2C6=9
Scan the empty cells in this row where 9 could go. Across the whole row, 9 has only one possible spot left. So R2C6 = 9.
14 Hidden Single ✓ R3C1=7
Scan the empty cells in this column where 7 could go. Across the whole column, 7 has only one possible spot left. So R3C1 = 7.
15 Hidden Single ✓ R3C3=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R3C3 = 1.
16 Hidden Single ✓ R3C8=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R3C8 = 4.
17 Hidden Single ✓ R4C5=9
Scan the empty cells in this column where 9 could go. Across the whole column, 9 has only one possible spot left. So R4C5 = 9.
18 Hidden Single ✓ R7C1=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R7C1 = 4.
19 Hidden Single ✓ R7C2=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R7C2 = 1.
20 Hidden Single ✓ R8C2=7
Scan the empty cells in this row where 7 could go. Across the whole row, 7 has only one possible spot left. So R8C2 = 7.
21 Naked Pair ✗ 4 eliminations
(1) Spot R3C4 and R4C4 on this column. (2) Spot R8C6 and R9C5 in this box. (1) These two cells can only hold 6 and 8: they reserve those digits. (2) These two cells can only hold 4 and 5: they reserve those digits. (1) 6 and 8 are therefore removed from the other cells of the column. (2) 4 and 5 are therefore removed from the other cells of the box.
22 Pointing ✗ 2 eliminations
Look at where 6 can go in this box. All its spots lie on a single column (R1C8 and R2C8). 6 is therefore removed from that column outside the box.
23 Two-String Kite ✗ 1 elimination
Find 4 on one row and one column, each with two candidate positions, sharing a box. The two strands meet in the shared box. One of the two free ends must hold 4. R5C9 is seen by both free ends: 4 is therefore eliminated there.
24 Two-String Kite ✗ 1 elimination
Find 5 on one row and one column, each with two candidate positions, sharing a box. The two strands meet in the shared box. One of the two free ends must hold 5. R2C7 is seen by both free ends: 5 is therefore eliminated there.
25 Naked Single ✓ R2C7=3
Look at cell R2C7. Every other digit already appears in its row, column or box. Only one option remains: R2C7 = 3.
26 Naked Pair ✗ 3 eliminations
Spot R1C8 and R2C8 on this column. These two cells can only hold 5 and 6: they reserve those digits. 5 and 6 are therefore removed from the other cells of the column.
27 Two-String Kite ✗ 1 elimination
Find 8 on one row and one column, each with two candidate positions, sharing a box. The two strands meet in the shared box. One of the two free ends must hold 8. R5C7 is seen by both free ends: 8 is therefore eliminated there.
28 Naked Single ✓ R5C7=4
Look at cell R5C7. Every other digit already appears in its row, column or box. Only one option remains: R5C7 = 4.
29 Hidden Single ✓ R3C4=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R3C4 = 6.
30 Hidden Single ✓ R3C5=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R3C5 = 5.
31 Hidden Single ✓ R3C6=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R3C6 = 8.
32 Hidden Single ✓ R4C4=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R4C4 = 8.
33 Hidden Single ✓ R4C6=4
Scan the empty cells in this row where 4 could go. Across the whole row, 4 has only one possible spot left. So R4C6 = 4.
34 Hidden Single ✓ R6C4=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R6C4 = 2.
35 Hidden Single ✓ R6C6=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R6C6 = 7.
36 Hidden Single ✓ R6C8=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R6C8 = 8.
37 Hidden Single ✓ R8C6=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R8C6 = 5.
38 Hidden Single ✓ R8C9=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R8C9 = 4.
39 Hidden Single ✓ R9C5=4
Scan the empty cells in this column where 4 could go. Across the whole column, 4 has only one possible spot left. So R9C5 = 4.
40 Hidden Single ✓ R9C6=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R9C6 = 2.
41 Naked Single ✓ R5C6=1
Look at cell R5C6. Every other digit already appears in its row, column or box. Only one option remains: R5C6 = 1.
42 Naked Single ✓ R6C1=5
Look at cell R6C1. Every other digit already appears in its row, column or box. Only one option remains: R6C1 = 5.
43 Naked Single ✓ R9C4=1
Look at cell R9C4. Every other digit already appears in its row, column or box. Only one option remains: R9C4 = 1.
44 Naked Single ✓ R1C1=8
Look at cell R1C1. Every other digit already appears in its row, column or box. Only one option remains: R1C1 = 8.
45 Hidden Single ✓ R1C3=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R1C3 = 5.
46 Hidden Single ✓ R1C8=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R1C8 = 6.
47 Hidden Single ✓ R2C2=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R2C2 = 6.
48 Hidden Single ✓ R2C8=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R2C8 = 5.
49 Hidden Single ✓ R4C7=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R4C7 = 7.
50 Hidden Single ✓ R4C9=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R4C9 = 5.
51 Hidden Single ✓ R5C2=8
Scan the empty cells in this row where 8 could go. Across the whole row, 8 has only one possible spot left. So R5C2 = 8.
52 Hidden Single ✓ R5C9=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R5C9 = 6.
53 Hidden Single ✓ R6C9=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R6C9 = 1.
54 Hidden Single ✓ R7C3=8
Scan the empty cells in this column where 8 could go. Across the whole column, 8 has only one possible spot left. So R7C3 = 8.
55 Hidden Single ✓ R7C7=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R7C7 = 5.
56 Hidden Single ✓ R9C2=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R9C2 = 5.
57 Hidden Single ✓ R9C7=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R9C7 = 8.
58 Hidden Single ✓ R9C8=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R9C8 = 7.
59 Hidden Single ✓ R9C9=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R9C9 = 3.
60 Naked Single ✓ R4C8=3
Look at cell R4C8. Every other digit already appears in its row, column or box. Only one option remains: R4C8 = 3.
61 Naked Single ✓ R5C5=3
Look at cell R5C5. Every other digit already appears in its row, column or box. Only one option remains: R5C5 = 3.
62 Naked Single ✓ R6C5=6
Look at cell R6C5. Every other digit already appears in its row, column or box. Only one option remains: R6C5 = 6.
63 Naked Single ✓ R4C3=6
Look at cell R4C3. Every other digit already appears in its row, column or box. Only one option remains: R4C3 = 6.
64 Naked Single ✓ R6C3=3
Look at cell R6C3. Every other digit already appears in its row, column or box. Only one option remains: R6C3 = 3.