LinkedIn Queens #916 Answer & Hints – November 2, 2026
Independent walkthrough by Queens Gamers · · 7×7 · Play on LinkedIn
LinkedIn Queens hints and the full solution for puzzle #916 on November 2, 2026. Follow 11 verified deductions on this 7×7 board, starting with region 3.
Region 1 (4 cells) · Region 2 (30 cells) · Region 3 (1 cell) · Region 4 (4 cells) · Region 5 (4 cells) · Region 6 (4 cells) · Region 7 (2 cells)
LinkedIn Queens #916 hints
Open one hint at a time. Reveal the full answer only when you are ready.
Hint 1
Start with region 3. Its candidates are R1C7. Which placement is forced? Remember that each row, column and region needs exactly one Queen.
Starting board: no deductions applied yet.
1 cells
R1C7Hint 2
Observe: region 3 has only one remaining candidate: R1C7.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R1C7. The 13 cells marked × conflict with it by row, column, region or diagonal adjacency.
Board after step 1; earlier deductions are already applied.
1 cells
R1C714 cells
R1C1, R1C2, R1C3, R1C4, R1C5, R1C6, R1C7, R2C6, R2C7, R3C7, R4C7, R5C7, R6C7, R7C7Hint 3
Step 1: Observe: region 3 has only one remaining candidate: R1C7.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R1C7. The 13 cells marked × conflict with it by row, column, region or diagonal adjacency.
Board after step 1; earlier deductions are already applied.
1 cells
R1C714 cells
R1C1, R1C2, R1C3, R1C4, R1C5, R1C6, R1C7, R2C6, R2C7, R3C7, R4C7, R5C7, R6C7, R7C7Reveal answer
Queens #916 full solution
| Row | Column | Region |
|---|---|---|
| R1 | C7 | 3 |
| R2 | C2 | 1 |
| R3 | C4 | 4 |
| R4 | C6 | 5 |
| R5 | C3 | 2 |
| R6 | C5 | 6 |
| R7 | C1 | 7 |
How to solve LinkedIn Queens #916
Step-by-step walkthrough
The key insight
The opening deduction: Observe: region 3 has only one remaining candidate: R1C7.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R1C7. The 13 cells marked × conflict with it by row, column, region or diagonal adjacency.
Your solving route
- Steps 1–1: narrow the candidates until R1C7 is forced.
- Steps 2–11: 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 3 has only one remaining candidate: R1C7.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R1C7. The 13 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R1C714 cells
R1C1, R1C2, R1C3, R1C4, R1C5, R1C6, R1C7, R2C6, R2C7, R3C7, R4C7, R5C7, R6C7, R7C7Step 2 · Locked candidates
Observe: region 1 has 2 remaining candidates: R2C1, R2C2. All lie in row 2.
Rule: region 1 needs one Queen, so it will occupy row 2. row 2 can contain no second Queen outside region 1.
Therefore: mark R2C3, R2C4, R2C5 ×.
2 cells
R2C1, R2C23 cells
R2C3, R2C4, R2C5Step 3 · Locked candidates
Observe: region 4 has 2 remaining candidates: R3C3, R3C4. All lie in row 3.
Rule: region 4 needs one Queen, so it will occupy row 3. row 3 can contain no second Queen outside region 4.
Therefore: mark R3C1, R3C2, R3C5, R3C6 ×.
2 cells
R3C3, R3C44 cells
R3C1, R3C2, R3C5, R3C6Step 4 · Locked candidates
Observe: region 5 has 2 remaining candidates: R4C5, R4C6. All lie in row 4.
Rule: region 5 needs one Queen, so it will occupy row 4. row 4 can contain no second Queen outside region 5.
Therefore: mark R4C1, R4C2, R4C3, R4C4 ×.
2 cells
R4C5, R4C64 cells
R4C1, R4C2, R4C3, R4C4Step 5 · Locked candidates
Observe: region 7 has 2 remaining candidates: R6C1, R7C1. All lie in column 1.
Rule: region 7 needs one Queen, so it will occupy column 1. column 1 can contain no second Queen outside region 7.
Therefore: mark R2C1, R5C1 ×.
2 cells
R6C1, R7C12 cells
R2C1, R5C1Step 6 · Only candidate
Observe: row 2 has only one remaining candidate: R2C2.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R2C2. The 4 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R2C25 cells
R2C2, R3C3, R5C2, R6C2, R7C2Step 7 · Only candidate
Observe: row 3 has only one remaining candidate: R3C4.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R3C4. The 4 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R3C45 cells
R3C4, R4C5, R5C4, R6C4, R7C4Step 8 · Only candidate
Observe: row 4 has only one remaining candidate: R4C6.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R4C6. The 4 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R4C65 cells
R4C6, R5C5, R5C6, R6C6, R7C6Step 9 · Only candidate
Observe: row 5 has only one remaining candidate: R5C3.
Rule: every row 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
R5C3, R6C3, R7C3, R7C5Step 10 · Only candidate
Observe: row 7 has only one remaining candidate: R7C1.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R7C1. The 1 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R7C12 cells
R6C1, R7C1Step 11 · Only candidate
Observe: row 6 has only one remaining candidate: R6C5.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R6C5. The 0 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R6C51 cells
R6C5A tempting mistake
A Queen at R1C1 looks possible before deductions. Following the forced placements leaves region 3 without a possible Queen. Rule out this tempting assumption.
Check every branch
Region 3 has no possible Queen: contradiction.