Sudoku Solution No. 115

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 (✗ RlCc≠d)
1 Hidden Single ✓ R1C6=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R1C6 = 5.
2 Hidden Single ✓ R2C2=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R2C2 = 3.
3 Hidden Single ✓ R2C4=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R2C4 = 9.
4 Hidden Single ✓ R2C9=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R2C9 = 5.
5 Hidden Single ✓ R3C3=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R3C3 = 9.
6 Hidden Single ✓ R3C6=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R3C6 = 2.
7 Hidden Single ✓ R5C4=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R5C4 = 5.
8 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.
9 Hidden Single ✓ R7C2=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R7C2 = 2.
10 Hidden Single ✓ R9C7=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R9C7 = 3.
11 Hidden Single ✓ R9C8=5
Scan the empty cells in this column where 5 could go. Across the whole column, 5 has only one possible spot left. So R9C8 = 5.
12 Naked Single ✓ R7C4=8
Look at cell R7C4. Every other digit already appears in its row, column or box. Only one option remains: R7C4 = 8.
13 Naked Single ✓ R7C8=1
Look at cell R7C8. Every other digit already appears in its row, column or box. Only one option remains: R7C8 = 1.
14 Hidden Single ✓ R1C5=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R1C5 = 8.
15 Hidden Single ✓ R3C7=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R3C7 = 1.
16 Hidden Single ✓ R3C8=8
Scan the empty cells in this column where 8 could go. Across the whole column, 8 has only one possible spot left. So R3C8 = 8.
17 Hidden Single ✓ R5C6=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R5C6 = 8.
18 Hidden Single ✓ R6C1=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R6C1 = 8.
19 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.
20 Hidden Single ✓ R8C7=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R8C7 = 8.
21 Hidden Single ✓ R9C2=8
Scan the empty cells in this row where 8 could go. Across the whole row, 8 has only one possible spot left. So R9C2 = 8.
22 Hidden Single ✓ R9C6=1
Scan the empty cells in this column where 1 could go. Across the whole column, 1 has only one possible spot left. So R9C6 = 1.
23 Naked Single ✓ R3C5=7
Look at cell R3C5. Every other digit already appears in its row, column or box. Only one option remains: R3C5 = 7.
24 Naked Single ✓ R3C1=6
Look at cell R3C1. Every other digit already appears in its row, column or box. Only one option remains: R3C1 = 6.
25 Naked Pair ✗ 5 eliminations
(1) Spot R5C8 and R6C8 in this box. (2) Spot R8C4 and R9C4 in this box. (1) These two cells can only hold 6 and 7: they reserve those digits. (2) These two cells can only hold 6 and 7: they reserve those digits. (1) 6 and 7 are therefore removed from the other cells of the box. (2) 6 and 7 are therefore removed from the other cells of the box.
26 Hidden Single ✓ R1C7=6
Scan the empty cells in this column where 6 could go. Across the whole column, 6 has only one possible spot left. So R1C7 = 6.
27 Hidden Single ✓ R1C9=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R1C9 = 2.
28 Hidden Single ✓ R2C7=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R2C7 = 7.
29 Naked Single ✓ R2C1=2
Look at cell R2C1. Every other digit already appears in its row, column or box. Only one option remains: R2C1 = 2.
30 Naked Pair ✗ 4 eliminations
Spot R4C5 and R8C5 on this column. These two cells can only hold 4 and 9: they reserve those digits. 4 and 9 are therefore removed from the other cells of the column.
31 Skyscraper ✗ 1 elimination
Find 9 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 9. R4C7 is seen by both free ends: 9 is therefore eliminated there.
32 Pointing ✗ 1 elimination
Look at where 9 can go in this box. All its spots lie on a single row (R5C7 and R5C9). 9 is therefore removed from that row outside the box.
33 Hidden Single ✓ R6C2=9
Scan the empty cells in this column where 9 could go. Across the whole column, 9 has only one possible spot left. So R6C2 = 9.
34 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. R8C3 is seen by both free ends: 4 is therefore eliminated there.
35 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. R4C7 is seen by both free ends: 4 is therefore eliminated there.
36 Naked Single ✓ R4C7=2
Look at cell R4C7. Every other digit already appears in its row, column or box. Only one option remains: R4C7 = 2.
37 Hidden Single ✓ R1C1=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R1C1 = 7.
38 Hidden Single ✓ R1C2=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R1C2 = 4.
39 Hidden Single ✓ R1C3=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R1C3 = 1.
40 Hidden Single ✓ R4C1=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R4C1 = 4.
41 Hidden Single ✓ R4C3=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R4C3 = 5.
42 Hidden Single ✓ R4C5=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R4C5 = 9.
43 Hidden Single ✓ R4C6=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R4C6 = 7.
44 Hidden Single ✓ R5C2=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R5C2 = 7.
45 Hidden Single ✓ R5C3=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R5C3 = 2.
46 Hidden Single ✓ R5C5=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R5C5 = 3.
47 Hidden Single ✓ R5C7=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R5C7 = 9.
48 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.
49 Hidden Single ✓ R5C9=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R5C9 = 4.
50 Hidden Single ✓ R6C3=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R6C3 = 3.
51 Hidden Single ✓ R6C5=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R6C5 = 6.
52 Hidden Single ✓ R6C6=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R6C6 = 4.
53 Hidden Single ✓ R6C8=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R6C8 = 7.
54 Hidden Single ✓ R7C6=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R7C6 = 9.
55 Hidden Single ✓ R7C7=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R7C7 = 4.
56 Hidden Single ✓ R8C1=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R8C1 = 5.
57 Hidden Single ✓ R8C2=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R8C2 = 1.
58 Hidden Single ✓ R8C3=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R8C3 = 7.
59 Hidden Single ✓ R8C4=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R8C4 = 6.
60 Hidden Single ✓ R8C5=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R8C5 = 4.
61 Hidden Single ✓ R8C9=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R8C9 = 9.
62 Hidden Single ✓ R9C3=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R9C3 = 4.
63 Hidden Single ✓ R9C4=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R9C4 = 7.
64 Hidden Single ✓ R9C9=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R9C9 = 6.