Sudoku Solution No. 62

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 ✓ R1C1=7
Scan the empty cells in this column where 7 could go. Across the whole column, 7 has only one possible spot left. So R1C1 = 7.
2 Hidden Single ✓ R1C9=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R1C9 = 4.
3 Hidden Single ✓ R2C7=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R2C7 = 3.
4 Hidden Single ✓ R2C8=1
Scan the empty cells in this column where 1 could go. Across the whole column, 1 has only one possible spot left. So R2C8 = 1.
5 Hidden Single ✓ R3C1=3
Scan the empty cells in this column where 3 could go. Across the whole column, 3 has only one possible spot left. So R3C1 = 3.
6 Hidden Single ✓ R7C7=7
Scan the empty cells in this row where 7 could go. Across the whole row, 7 has only one possible spot left. So R7C7 = 7.
7 Hidden Single ✓ R9C8=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R9C8 = 2.
8 Hidden Single ✓ R4C9=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R4C9 = 3.
9 Hidden Single ✓ R5C6=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R5C6 = 3.
10 Hidden Single ✓ R6C8=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R6C8 = 5.
11 Hidden Single ✓ R6C9=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R6C9 = 7.
12 Hidden Single ✓ R7C8=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R7C8 = 3.
13 Hidden Single ✓ R8C5=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R8C5 = 3.
14 Hidden Single ✓ R4C6=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R4C6 = 9.
15 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.
16 Hidden Single ✓ R8C2=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R8C2 = 9.
17 Hidden Single ✓ R8C6=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R8C6 = 1.
18 Hidden Single ✓ R9C9=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R9C9 = 9.
19 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.
20 Naked Single ✓ R9C5=4
Look at cell R9C5. Every other digit already appears in its row, column or box. Only one option remains: R9C5 = 4.
21 Naked Single ✓ R7C1=5
Look at cell R7C1. Every other digit already appears in its row, column or box. Only one option remains: R7C1 = 5.
22 Naked Single ✓ R7C4=9
Look at cell R7C4. Every other digit already appears in its row, column or box. Only one option remains: R7C4 = 9.
23 Naked Single ✓ R9C4=5
Look at cell R9C4. Every other digit already appears in its row, column or box. Only one option remains: R9C4 = 5.
24 Hidden Single ✓ R2C2=4
Scan the empty cells in this row where 4 could go. Across the whole row, 4 has only one possible spot left. So R2C2 = 4.
25 Hidden Single ✓ R2C6=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R2C6 = 5.
26 Hidden Single ✓ R3C5=9
Scan the empty cells in this row where 9 could go. Across the whole row, 9 has only one possible spot left. So R3C5 = 9.
27 Hidden Single ✓ R3C9=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R3C9 = 8.
28 Hidden Single ✓ R8C9=5
Scan the empty cells in this row where 5 could go. Across the whole row, 5 has only one possible spot left. So R8C9 = 5.
29 Naked Pair ✗ 11 eliminations
(1) Spot R4C2 and R4C7 on this row. (2) Spot R8C1 and R9C1 on this column. (1) These two cells can only hold 6 and 8: they reserve those digits. (2) These two cells can only hold 6 and 8: they reserve those digits. (1) 6 and 8 are therefore removed from the other cells of the row. (2) 6 and 8 are therefore removed from the other cells of the column.
30 Naked Single ✓ R4C8=4
Look at cell R4C8. Every other digit already appears in its row, column or box. Only one option remains: R4C8 = 4.
31 Naked Single ✓ R5C1=4
Look at cell R5C1. Every other digit already appears in its row, column or box. Only one option remains: R5C1 = 4.
32 Hidden Single ✓ R6C4=4
Scan the empty cells in this row where 4 could go. Across the whole row, 4 has only one possible spot left. So R6C4 = 4.
33 Naked Pair ✗ 2 eliminations
Spot R4C1 and R6C1 in this box. These two cells can only hold 1 and 2: they reserve those digits. 1 and 2 are therefore removed from the other cells of the box.
34 Two-String Kite ✗ 1 elimination
Find 6 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 6. R3C3 is seen by both free ends: 6 is therefore eliminated there.
35 X-Wing ✗ 1 elimination
Find where 6 appears on two columns: altogether, only two rows are involved (R1C3, R6C3, R1C6 and R6C6). On each covered row, 6 must lie within one of these two columns. 6 is therefore removed from those two rows outside the two X-Wing columns.
36 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. R1C5 is seen by both free ends: 8 is therefore eliminated there.
37 Hidden Pair ✗ 1 elimination
Find where 7 and 8 can go in this box. These two digits only fit in R2C4 and R2C5 in the box. The other candidates in R2C4 and R2C5 are therefore eliminated.
38 Hidden Single ✓ R1C2=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R1C2 = 8.
39 Hidden Single ✓ R1C3=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R1C3 = 6.
40 Hidden Single ✓ R1C5=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R1C5 = 1.
41 Hidden Single ✓ R1C6=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R1C6 = 2.
42 Hidden Single ✓ R1C7=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R1C7 = 5.
43 Hidden Single ✓ R2C3=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R2C3 = 2.
44 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.
45 Hidden Single ✓ R2C5=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R2C5 = 8.
46 Hidden Single ✓ R3C2=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R3C2 = 5.
47 Hidden Single ✓ R3C3=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R3C3 = 1.
48 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.
49 Hidden Single ✓ R3C7=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R3C7 = 2.
50 Hidden Single ✓ R4C1=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R4C1 = 2.
51 Hidden Single ✓ R4C2=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R4C2 = 6.
52 Hidden Single ✓ R4C4=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R4C4 = 1.
53 Hidden Single ✓ R4C5=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R4C5 = 7.
54 Hidden Single ✓ R4C7=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R4C7 = 8.
55 Hidden Single ✓ R5C4=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R5C4 = 8.
56 Hidden Single ✓ R5C8=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R5C8 = 6.
57 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.
58 Hidden Single ✓ R6C3=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R6C3 = 8.
59 Hidden Single ✓ R6C5=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R6C5 = 2.
60 Hidden Single ✓ R6C6=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R6C6 = 6.
61 Hidden Single ✓ R8C1=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R8C1 = 6.
62 Hidden Single ✓ R8C8=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R8C8 = 8.
63 Hidden Single ✓ R9C1=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R9C1 = 8.
64 Hidden Single ✓ R9C7=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R9C7 = 6.