Pips answers for Saturday, February 21, 2026

Start with the 9 region at the left: the only way to make it with these dominoes is 3 + 6.

• the single-cell 0 region at the top-left: 0
• the single-cell 5 region at the top-right: 5
• the 9 region at the left: 3, 6
• the 9 region at the right: 1, 2, 6
• the single-cell 5 region at the bottom-left: 5
• the single-cell 0 region at the bottom-right: 0

1. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the single-cell 0 region at the top-left needs a total of exactly 0 — nothing else fits. Place the 0–6 domino vertically on row 1, column 1 and row 2, column 1.
2. The 9 region at the left needs a total of exactly 9 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–5 domino vertically on row 2, column 2 and row 3, column 2.
3. Row 1, column 5 has only one open neighbor, so its domino must run to row 2, column 5; the single-cell 5 region at the top-right needs a total of exactly 5 — nothing else fits. Place the 2–5 domino vertically on row 1, column 5 and row 2, column 5.
4. The 9 region at the right needs a total of exactly 9 (1 of its 3 cells is already filled) — nothing else fits. Place the 0–1 domino vertically on row 2, column 4 and row 3, column 4.
5. Row 2, column 3 has only one open neighbor, so its domino must run to row 1, column 3; the 9 region at the right needs a total of exactly 9 (2 of its 3 cells are already filled) — nothing else fits. Place the 3–6 domino vertically on row 1, column 3 and row 2, column 3.

Start with the 10 region at the bottom-right: the only way to make it with these dominoes is 4 + 6.

• the single-cell “>3” region at the top: 6
• the single-cell “>3” region at the top: 5
• the 2 region at the center: 1, 1
• the 4 region at the center: 2, 2
• the 2 region at the bottom-left: 1, 1
• the 10 region at the bottom-right: 4, 6
• the 7 region at the bottom-left: 3, 4

1. The 10 region at the bottom-right needs a total of exactly 10 — 5 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–4 domino vertically on row 2, column 6 and row 3, column 6.
2. Row 4, column 6 has only one open neighbor, so its domino must run to row 4, column 5; the 10 region at the bottom-right needs a total of exactly 10 (1 of its 2 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–6 domino horizontally on row 4, column 5 and row 4, column 6.
3. The single-cell “>3” region at the top must total more than 3 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–6 domino horizontally on row 1, column 3 and row 1, column 4.
4. The 2 region at the bottom-left needs a total of exactly 2 — 6 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–4 domino horizontally on row 4, column 1 and row 4, column 2.
5. The 7 region at the bottom-left needs a total of exactly 7 (1 of its 2 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–3 domino vertically on row 3, column 3 and row 4, column 3.
6. Row 2, column 3 has only one open neighbor, so its domino must run to row 2, column 4; the 2 region at the center needs a total of exactly 2 (1 of its 2 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–2 domino horizontally on row 2, column 3 and row 2, column 4.
7. Row 3, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 2 region at the bottom-left needs a total of exactly 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–1 domino vertically on row 2, column 1 and row 3, column 1.
8. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 4 region at the center needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–2 domino vertically on row 3, column 4 and row 4, column 4.

This board has more than one valid layout; the steps follow the official answer.

Start with the 0 region at the top-left: the only way to make it with these dominoes is 0 + 0.

• the single-cell 0 region at the top-left: 0
• the 0 region at the top-left: 0, 0
• the 7 region at the top: 0, 2, 2, 3
• the “=” region at the top-right: 5, 5, 5, 5, 5
• the “=” region at the right: 6, 6, 6
• the “=” region at the left: 4, 4
• the “=” region at the bottom: 1, 1, 1
• the 7 region at the right: 3, 4
• the 7 region at the bottom-left: 1, 1, 5
• the 7 region at the bottom: 2, 5
• the single-cell 0 region at the bottom-right: 0

1. The 0 region at the top-left needs a total of exactly 0 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–0 domino horizontally on row 1, column 1 and row 1, column 2.
2. The 7 region at the right needs a total of exactly 7 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–3 domino vertically on row 4, column 8 and row 5, column 8.
3. The “=” region at the top-right needs every pip to match — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–5 domino horizontally on row 1, column 7 and row 1, column 8.
4. Row 2, column 8 has only one open neighbor, so its domino must run to row 3, column 8; the “=” region at the top-right needs every pip to match (2 of its 5 cells are already filled) — nothing else fits. Place the 4–5 domino vertically on row 2, column 8 and row 3, column 8.
5. The “=” region at the top-right needs every pip to match (3 of its 5 cells are already filled) — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–5 domino horizontally on row 1, column 4 and row 1, column 5.
6. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the 0 region at the top-left needs a total of exactly 0 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–2 domino vertically on row 1, column 3 and row 2, column 3.
7. The “=” region at the right needs every pip to match — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 6–6 domino vertically on row 3, column 6 and row 4, column 6.
8. Row 1, column 6 has only one open neighbor, so its domino must run to row 2, column 6; the “=” region at the top-right needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 5–6 domino vertically on row 1, column 6 and row 2, column 6.
9. The 7 region at the bottom needs a total of exactly 7 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–5 domino horizontally on row 5, column 5 and row 5, column 6.
10. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the 7 region at the top needs a total of exactly 7 (2 of its 4 cells are already filled) — 10 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–3 domino vertically on row 2, column 4 and row 3, column 4.
11. Row 3, column 3 has only one open neighbor, so its domino must run to row 4, column 3; the 7 region at the top needs a total of exactly 7 (3 of its 4 cells are already filled) — nothing else fits. Place the 2–3 domino vertically on row 3, column 3 and row 4, column 3.
12. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the “=” region at the bottom needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 1–1 domino vertically on row 4, column 4 and row 5, column 4.
13. The 7 region at the bottom-left needs a total of exactly 7 — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–5 domino horizontally on row 5, column 2 and row 5, column 3.
14. The “=” region at the left needs every pip to match — nothing else fits. Place the 1–4 domino vertically on row 4, column 1 and row 5, column 1.
15. Row 3, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the “=” region at the left needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino vertically on row 2, column 1 and row 3, column 1.

This board has more than one valid layout; the steps follow the official answer.