LinkedIn Queens #891 Answer & Hints – October 8, 2026
Independent walkthrough by Queens Gamers · · 8×8 · Play on LinkedIn
LinkedIn Queens hints and the full solution for puzzle #891 on October 8, 2026. Follow 15 verified deductions on this 8×8 board, starting with region 3 in row 1.
Region 1 (4 cells) · Region 2 (20 cells) · Region 3 (2 cells) · Region 4 (20 cells) · Region 5 (5 cells) · Region 6 (8 cells) · Region 7 (3 cells) · Region 8 (2 cells)
LinkedIn Queens #891 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, R1C8. Can you spot a shared row, column or region? Remember that each row, column and region needs exactly one Queen.
Starting board: no deductions applied yet.
2 cells
R1C7, R1C8Hint 2
Observe: region 3 has 2 remaining candidates: R1C7, R1C8. All lie in row 1.
Rule: region 3 needs one Queen, so it will occupy row 1. row 1 can contain no second Queen outside region 3.
Therefore: mark R1C1, R1C2, R1C3, R1C4, R1C5, R1C6 ×.
Board after step 1; earlier deductions are already applied.
2 cells
R1C7, R1C86 cells
R1C1, R1C2, R1C3, R1C4, R1C5, R1C6Hint 3
Step 1: Observe: region 3 has 2 remaining candidates: R1C7, R1C8. All lie in row 1.
Rule: region 3 needs one Queen, so it will occupy row 1. row 1 can contain no second Queen outside region 3.
Therefore: mark R1C1, R1C2, R1C3, R1C4, R1C5, R1C6 ×.
Step 2: Observe: region 1 has 2 remaining candidates: R2C2, R3C2. All lie in column 2.
Rule: region 1 needs one Queen, so it will occupy column 2. column 2 can contain no second Queen outside region 1.
Therefore: mark R4C2, R5C2, R6C2, R7C2, R8C2 ×.
Step 3: Observe: region 8 has only one remaining candidate: R8C1.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R8C1. The 12 cells marked × conflict with it by row, column, region or diagonal adjacency.
Board after step 3; earlier deductions are already applied.
1 cells
R8C113 cells
R2C1, R3C1, R4C1, R5C1, R6C1, R7C1, R8C1, R8C3, R8C4, R8C5, R8C6, R8C7, R8C8Reveal answer
Queens #891 full solution
| Row | Column | Region |
|---|---|---|
| R1 | C8 | 3 |
| R2 | C2 | 1 |
| R3 | C5 | 2 |
| R4 | C3 | 5 |
| R5 | C6 | 6 |
| R6 | C4 | 4 |
| R7 | C7 | 7 |
| R8 | C1 | 8 |
How to solve LinkedIn Queens #891
Step-by-step walkthrough
The key insight
The opening deduction: Observe: region 3 has 2 remaining candidates: R1C7, R1C8. All lie in row 1.
Rule: region 3 needs one Queen, so it will occupy row 1. row 1 can contain no second Queen outside region 3.
Therefore: mark R1C1, R1C2, R1C3, R1C4, R1C5, R1C6 ×.
Your solving route
- Steps 1–3: narrow the candidates until R8C1 is forced.
- Steps 4–15: 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 · Locked candidates
Observe: region 3 has 2 remaining candidates: R1C7, R1C8. All lie in row 1.
Rule: region 3 needs one Queen, so it will occupy row 1. row 1 can contain no second Queen outside region 3.
Therefore: mark R1C1, R1C2, R1C3, R1C4, R1C5, R1C6 ×.
2 cells
R1C7, R1C86 cells
R1C1, R1C2, R1C3, R1C4, R1C5, R1C6Step 2 · Locked candidates
Observe: region 1 has 2 remaining candidates: R2C2, R3C2. All lie in column 2.
Rule: region 1 needs one Queen, so it will occupy column 2. column 2 can contain no second Queen outside region 1.
Therefore: mark R4C2, R5C2, R6C2, R7C2, R8C2 ×.
2 cells
R2C2, R3C25 cells
R4C2, R5C2, R6C2, R7C2, R8C2Step 3 · Only candidate
Observe: region 8 has only one remaining candidate: R8C1.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R8C1. The 12 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R8C113 cells
R2C1, R3C1, R4C1, R5C1, R6C1, R7C1, R8C1, R8C3, R8C4, R8C5, R8C6, R8C7, R8C8Step 4 · Only candidate
Observe: region 7 has only one remaining candidate: R7C7.
Rule: every region needs exactly one Queen.
Therefore: place a Queen at R7C7. The 13 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R7C714 cells
R1C7, R2C7, R3C7, R4C7, R5C7, R6C6, R6C7, R6C8, R7C3, R7C4, R7C5, R7C6, R7C7, R7C8Step 5 · Only candidate
Observe: row 1 has only one remaining candidate: R1C8.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R1C8. The 4 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R1C85 cells
R1C8, R2C8, R3C8, R4C8, R5C8Step 6 · Locked candidates
Observe: region 4 has 2 remaining candidates: R6C3, R6C4. All lie in row 6.
Rule: region 4 needs one Queen, so it will occupy row 6. row 6 can contain no second Queen outside region 4.
Therefore: mark R6C5 ×.
2 cells
R6C3, R6C41 cells
R6C5Step 7 · Common exclusions
Observe: row 3 must contain one Queen, chosen from R3C2, R3C3, R3C4, R3C5, R3C6.
Rule: Queens cannot share a row, column or region, or touch diagonally. R2C3 conflicts with R3C2, R3C4 through diagonally adjacent cells; with R3C3 through the same column (column 3); with R3C5, R3C6 through the same region (region 2).
Therefore: mark R2C3 ×. Every possible Queen in row 3 rules out these cells.
5 cells
R3C2, R3C3, R3C4, R3C5, R3C61 cells
R2C3Step 8 · Common exclusions
Observe: row 5 must contain one Queen, chosen from R5C3, R5C4, R5C5, R5C6.
Rule: Queens cannot share a row, column or region, or touch diagonally. R4C4 conflicts with R5C3 through diagonally adjacent cells; with R5C4 through the same column (column 4); with R5C5, R5C6 through the same region (region 6).
Therefore: mark R4C4 ×. Every possible Queen in row 5 rules out these cells.
4 cells
R5C3, R5C4, R5C5, R5C61 cells
R4C4Step 9 · Common exclusions
Observe: row 6 must contain one Queen, chosen from R6C3, R6C4.
Rule: Queens cannot share a row, column or region, or touch diagonally. R5C3 conflicts with R6C3 through the same column (column 3); with R6C4 through diagonally adjacent cells. R5C4 conflicts with R6C3 through diagonally adjacent cells; with R6C4 through the same column (column 4).
Therefore: mark R5C3, R5C4 ×. Every possible Queen in row 6 rules out these cells.
2 cells
R6C3, R6C42 cells
R5C3, R5C4Step 10 · Locked candidates
Observe: row 5 has 2 remaining candidates: R5C5, R5C6. All lie in region 6.
Rule: row 5 needs one Queen, so it will occupy region 6. region 6 can contain no second Queen outside row 5.
Therefore: mark R4C5, R4C6 ×.
2 cells
R5C5, R5C62 cells
R4C5, R4C6Step 11 · Only candidate
Observe: row 4 has only one remaining candidate: R4C3.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R4C3. The 4 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R4C35 cells
R3C2, R3C3, R3C4, R4C3, R6C3Step 12 · Only candidate
Observe: row 6 has only one remaining candidate: R6C4.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R6C4. The 2 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R6C43 cells
R2C4, R5C5, R6C4Step 13 · Only candidate
Observe: row 5 has only one remaining candidate: R5C6.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R5C6. The 2 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R5C63 cells
R2C6, R3C6, R5C6Step 14 · Only candidate
Observe: row 3 has only one remaining candidate: R3C5.
Rule: every row needs exactly one Queen.
Therefore: place a Queen at R3C5. The 1 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R3C52 cells
R2C5, R3C5Step 15 · 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 0 cells marked × conflict with it by row, column, region or diagonal adjacency.
1 cells
R2C21 cells
R2C2A 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.