LinkedIn Queens #902 Answer & Hints – October 19, 2026
Independent walkthrough by Queens Gamers · · 7×7 · Play on LinkedIn
LinkedIn Queens hints and the full solution for puzzle #902 on October 19, 2026. Follow 10 verified deductions on this 7×7 board, starting with region 4.
Region 1 (10 cells) · Region 2 (14 cells) · Region 3 (8 cells) · Region 4 (1 cell) · Region 5 (7 cells) · Region 6 (5 cells) · Region 7 (4 cells)
LinkedIn Queens #902 hints
Open one hint at a time. Reveal the full answer only when you are ready.
Hint 1
Start with region 4. Its candidates are R2C4. Which placement is forced? Remember that each row, column and region needs exactly one Queen.
Starting board: no deductions applied yet.
1 cells
R2C4Hint 2
Observe: region 4 has only one remaining candidate: R2C4.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R2C4. The 16 cells marked × conflict with it by row, column, region or diagonal adjacency.
Board after step 1; earlier deductions are already applied.
1 cells
R2C417 cells
R1C3, R1C4, R1C5, R2C1, R2C2, R2C3, R2C4, R2C5, R2C6, R2C7, R3C3, R3C4, R3C5, R4C4, R5C4, R6C4, R7C4Hint 3
Step 1: Observe: region 4 has only one remaining candidate: R2C4.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R2C4. The 16 cells marked × conflict with it by row, column, region or diagonal adjacency.
Board after step 1; earlier deductions are already applied.
1 cells
R2C417 cells
R1C3, R1C4, R1C5, R2C1, R2C2, R2C3, R2C4, R2C5, R2C6, R2C7, R3C3, R3C4, R3C5, R4C4, R5C4, R6C4, R7C4Reveal answer
Queens #902 full solution
| Row | Column | Region |
|---|---|---|
| R1 | C6 | 3 |
| R2 | C4 | 4 |
| R3 | C1 | 1 |
| R4 | C5 | 6 |
| R5 | C3 | 2 |
| R6 | C7 | 7 |
| R7 | C2 | 5 |
How to solve LinkedIn Queens #902
Step-by-step walkthrough
The key insight
The opening deduction: Observe: region 4 has only one remaining candidate: R2C4.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R2C4. The 16 cells marked × conflict with it by row, column, region or diagonal adjacency.
Your solving route
- Steps 1–1: narrow the candidates until R2C4 is forced.
- Steps 2–10: use the confirmed Queens to eliminate conflicts, then repeat the deductions for the remaining regions.
Open each deduction to see the observation, rule and board changes.
Step 1 · Only candidate
Observe: region 4 has only one remaining candidate: R2C4.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R2C4. The 16 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R2C417 cells
R1C3, R1C4, R1C5, R2C1, R2C2, R2C3, R2C4, R2C5, R2C6, R2C7, R3C3, R3C4, R3C5, R4C4, R5C4, R6C4, R7C4Step 2 · Locked candidates
Observe: region 2 has 2 remaining candidates: R5C3, R5C5. All lie in row 5.
Rule: region 2 needs one Queen, so it will occupy row 5. row 5 can contain no second Queen outside region 2.
Therefore: mark R5C1, R5C2, R5C6, R5C7 ×.
2 cells
R5C3, R5C54 cells
R5C1, R5C2, R5C6, R5C7Step 3 · Locked candidates
Observe: column 1 has 5 remaining candidates: R1C1, R3C1, R4C1, R6C1, R7C1. All lie in region 1.
Rule: column 1 needs one Queen, so it will occupy region 1. region 1 can contain no second Queen outside column 1.
Therefore: mark R1C2, R3C2 ×.
5 cells
R1C1, R3C1, R4C1, R6C1, R7C12 cells
R1C2, R3C2Step 4 · Locked candidates
Observe: column 2 has 3 remaining candidates: R4C2, R6C2, R7C2. All lie in region 5.
Rule: column 2 needs one Queen, so it will occupy region 5. region 5 can contain no second Queen outside column 2.
Therefore: mark R4C3, R6C3, R7C3 ×.
3 cells
R4C2, R6C2, R7C23 cells
R4C3, R6C3, R7C3Step 5 · Only candidate
Observe: column 3 has only one remaining candidate: R5C3.
Rule: every column needs exactly one Queen.
Therefore: place a Queen at R5C3. The 3 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R5C34 cells
R4C2, R5C3, R5C5, R6C2Step 6 · Only candidate
Observe: column 2 has only one remaining candidate: R7C2.
Rule: every column needs exactly one Queen.
Therefore: place a Queen at R7C2. The 5 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R7C26 cells
R6C1, R7C1, R7C2, R7C5, R7C6, R7C7Step 7 · Only candidate
Observe: region 7 has only one remaining candidate: R6C7.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R6C7. The 5 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R6C76 cells
R1C7, R3C7, R4C7, R6C5, R6C6, R6C7Step 8 · Only candidate
Observe: column 5 has only one remaining candidate: R4C5.
Rule: every column needs exactly one Queen.
Therefore: place a Queen at R4C5. The 3 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R4C54 cells
R3C6, R4C1, R4C5, R4C6Step 9 · Only candidate
Observe: row 3 has only one remaining candidate: R3C1.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R3C1. The 1 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R3C12 cells
R1C1, R3C1Step 10 · Only candidate
Observe: row 1 has only one remaining candidate: R1C6.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R1C6. The 0 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R1C61 cells
R1C6A tempting mistake
A Queen at R1C1 looks possible before deductions. All 3 terminal branches run out of candidates in row 4, row 6, region 6. Rule out this tempting assumption.
Check every branch
The only candidate in region 4 forces R2C4.
Try R3C6.
The only candidate in region 6 forces R6C5.
The only candidate in column 7 forces R7C7.
The only candidate in region 2 forces R5C3.
Row 4 has no possible Queen: contradiction.
Try R3C7.
Try R5C3.
The only candidate in row 4 forces R4C5.
Row 6 has no possible Queen: contradiction.
Try R5C5.
Region 6 has no possible Queen: contradiction.