Sudoku Solution No. 121

Hard Score 72 / 100
Digits appear one by one.

Step-by-step solution

Steps 65 Estimated time 104 – 156 min Hardest technique X-Wing Clues 23
Given digit (clue) Solved digit Placement (✓ RlCc=d) Elimination (✗ RlCc≠d)
1 Naked Single ✓ R7C3=4
Look at cell R7C3. Every other digit already appears in its row, column or box. Only one option remains: R7C3 = 4.
2 Naked Single ✓ R9C3=3
Look at cell R9C3. Every other digit already appears in its row, column or box. Only one option remains: R9C3 = 3.
3 Naked Single ✓ R8C2=7
Look at cell R8C2. Every other digit already appears in its row, column or box. Only one option remains: R8C2 = 7.
4 Naked Single ✓ R8C3=6
Look at cell R8C3. Every other digit already appears in its row, column or box. Only one option remains: R8C3 = 6.
5 Naked Single ✓ R7C2=2
Look at cell R7C2. Every other digit already appears in its row, column or box. Only one option remains: R7C2 = 2.
6 Hidden Single ✓ R1C2=5
Scan the empty cells in this row where 5 could go. Across the whole row, 5 has only one possible spot left. So R1C2 = 5.
7 Hidden Single ✓ R1C8=6
Scan the empty cells in this column where 6 could go. Across the whole column, 6 has only one possible spot left. So R1C8 = 6.
8 Hidden Single ✓ R2C4=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R2C4 = 2.
9 Hidden Single ✓ R3C2=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R3C2 = 4.
10 Hidden Single ✓ R3C6=6
Scan the empty cells in this row where 6 could go. Across the whole row, 6 has only one possible spot left. So R3C6 = 6.
11 Hidden Single ✓ R4C6=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R4C6 = 2.
12 Hidden Single ✓ R5C1=2
Scan the empty cells in this column where 2 could go. Across the whole column, 2 has only one possible spot left. So R5C1 = 2.
13 Hidden Single ✓ R5C4=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R5C4 = 6.
14 Hidden Single ✓ R7C6=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R7C6 = 1.
15 Hidden Single ✓ R1C4=1
Scan the empty cells in this column where 1 could go. Across the whole column, 1 has only one possible spot left. So R1C4 = 1.
16 Hidden Single ✓ R2C2=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R2C2 = 9.
17 Hidden Single ✓ R2C3=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R2C3 = 1.
18 Hidden Single ✓ R3C7=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R3C7 = 1.
19 Hidden Single ✓ R3C9=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R3C9 = 2.
20 Hidden Single ✓ R4C4=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R4C4 = 9.
21 Hidden Single ✓ R5C3=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R5C3 = 9.
22 Hidden Single ✓ R6C1=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R6C1 = 1.
23 Hidden Single ✓ R6C3=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R6C3 = 5.
24 Hidden Single ✓ R7C8=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R7C8 = 5.
25 Hidden Single ✓ R8C7=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R8C7 = 9.
26 Hidden Single ✓ R8C9=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R8C9 = 3.
27 Hidden Single ✓ R9C6=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R9C6 = 9.
28 Hidden Single ✓ R9C7=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R9C7 = 2.
29 Naked Single ✓ R3C3=8
Look at cell R3C3. Every other digit already appears in its row, column or box. Only one option remains: R3C3 = 8.
30 Naked Single ✓ R8C4=4
Look at cell R8C4. Every other digit already appears in its row, column or box. Only one option remains: R8C4 = 4.
31 Naked Single ✓ R8C6=5
Look at cell R8C6. Every other digit already appears in its row, column or box. Only one option remains: R8C6 = 5.
32 Hidden Single ✓ R1C5=9
Scan the empty cells in this row where 9 could go. Across the whole row, 9 has only one possible spot left. So R1C5 = 9.
33 Hidden Single ✓ R4C9=5
Scan the empty cells in this row where 5 could go. Across the whole row, 5 has only one possible spot left. So R4C9 = 5.
34 Hidden Single ✓ R5C5=5
Scan the empty cells in this row where 5 could go. Across the whole row, 5 has only one possible spot left. So R5C5 = 5.
35 Hidden Single ✓ R6C5=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R6C5 = 4.
36 Pointing ✗ 1 elimination
Look at where 3 can go in this box. All its spots lie on a single column (R5C6 and R6C6). 3 is therefore removed from that column outside the box.
37 Two-String Kite ✗ 1 elimination
Find 7 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 7. R6C7 is seen by both free ends: 7 is therefore eliminated there.
38 Naked Pair ✗ 4 eliminations
Spot R4C8 and R6C7 in this box. These two cells can only hold 3 and 8: they reserve those digits. 3 and 8 are therefore removed from the other cells of the box.
39 Naked Pair ✗ 1 elimination
Spot R5C7 and R5C8 on this row. These two cells can only hold 4 and 7: they reserve those digits. 4 and 7 are therefore removed from the other cells of the row.
40 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. R6C7 is seen by both free ends: 8 is therefore eliminated there.
41 Naked Single ✓ R6C7=3
Look at cell R6C7. Every other digit already appears in its row, column or box. Only one option remains: R6C7 = 3.
42 Naked Single ✓ R4C8=8
Look at cell R4C8. Every other digit already appears in its row, column or box. Only one option remains: R4C8 = 8.
43 Naked Single ✓ R4C2=3
Look at cell R4C2. Every other digit already appears in its row, column or box. Only one option remains: R4C2 = 3.
44 Naked Single ✓ R5C2=8
Look at cell R5C2. Every other digit already appears in its row, column or box. Only one option remains: R5C2 = 8.
45 Naked Single ✓ R5C6=3
Look at cell R5C6. Every other digit already appears in its row, column or box. Only one option remains: R5C6 = 3.
46 Hidden Single ✓ R1C1=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R1C1 = 3.
47 Hidden Single ✓ R2C8=3
Scan the empty cells in this column where 3 could go. Across the whole column, 3 has only one possible spot left. So R2C8 = 3.
48 Hidden Single ✓ R3C1=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R3C1 = 7.
49 Hidden Single ✓ R3C5=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R3C5 = 3.
50 Skyscraper ✗ 2 eliminations
Find 7 on two rows, each with exactly two candidate positions. A shared column connects one end of each pair. Whichever way the logic falls, one of the two free ends must hold 7. R1C7 and R9C9 is seen by both free ends: 7 is therefore eliminated there.
51 Skyscraper ✗ 2 eliminations
Find 8 on two rows, each with exactly two candidate positions. A shared column connects one end of each pair. Whichever way the logic falls, one of the two free ends must hold 8. R1C7 and R9C9 is seen by both free ends: 8 is therefore eliminated there.
52 Naked Single ✓ R1C7=4
Look at cell R1C7. Every other digit already appears in its row, column or box. Only one option remains: R1C7 = 4.
53 Naked Single ✓ R5C7=7
Look at cell R5C7. Every other digit already appears in its row, column or box. Only one option remains: R5C7 = 7.
54 Naked Single ✓ R5C8=4
Look at cell R5C8. Every other digit already appears in its row, column or box. Only one option remains: R5C8 = 4.
55 Naked Single ✓ R7C7=8
Look at cell R7C7. Every other digit already appears in its row, column or box. Only one option remains: R7C7 = 8.
56 Naked Single ✓ R9C8=7
Look at cell R9C8. Every other digit already appears in its row, column or box. Only one option remains: R9C8 = 7.
57 Naked Single ✓ R9C9=4
Look at cell R9C9. Every other digit already appears in its row, column or box. Only one option remains: R9C9 = 4.
58 Naked Single ✓ R7C5=7
Look at cell R7C5. Every other digit already appears in its row, column or box. Only one option remains: R7C5 = 7.
59 Naked Single ✓ R9C4=8
Look at cell R9C4. Every other digit already appears in its row, column or box. Only one option remains: R9C4 = 8.
60 Naked Single ✓ R2C5=8
Look at cell R2C5. Every other digit already appears in its row, column or box. Only one option remains: R2C5 = 8.
61 Naked Single ✓ R2C9=7
Look at cell R2C9. Every other digit already appears in its row, column or box. Only one option remains: R2C9 = 7.
62 Naked Single ✓ R6C4=7
Look at cell R6C4. Every other digit already appears in its row, column or box. Only one option remains: R6C4 = 7.
63 Naked Single ✓ R6C6=8
Look at cell R6C6. Every other digit already appears in its row, column or box. Only one option remains: R6C6 = 8.
64 Naked Single ✓ R1C6=7
Look at cell R1C6. Every other digit already appears in its row, column or box. Only one option remains: R1C6 = 7.
65 Naked Single ✓ R1C9=8
Look at cell R1C9. Every other digit already appears in its row, column or box. Only one option remains: R1C9 = 8.