Pips answers for Thursday, August 21, 2025

Start with the 9 region at the bottom: the only way to make it with these dominoes is 4 + 5.

• the single-cell 2 region at the top-left: 2
• the 7 region at the top: 3, 4
• the 7 region at the center: 3, 4
• the 9 region at the bottom: 4, 5
• the single-cell 2 region at the bottom-right: 2

1. The 7 region at the top needs a total of exactly 7 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–3 domino horizontally on row 1, column 1 and row 1, column 2.
2. The 9 region at the bottom needs a total of exactly 9 — nothing else fits. Place the 2–4 domino horizontally on row 6, column 2 and row 6, column 3.
3. The 7 region at the center needs a total of exactly 7 — nothing else fits. Place the 4–5 domino vertically on row 4, column 2 and row 5, column 2.
4. Row 2, column 2 has only one open neighbor, so its domino must run to row 3, column 2; the 7 region at the top needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–4 domino vertically on row 2, column 2 and row 3, column 2.

Start with the 5 region at the bottom: the only way to make it with these dominoes is 0 + 1 + 1 + 1 + 2.

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

1. The 5 region at the bottom needs a total of exactly 5 — 2 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 4, column 3 and row 4, column 4.
2. The 5 region at the bottom needs a total of exactly 5 (1 of its 5 cells is already filled) — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–1 domino vertically on row 5, column 3 and row 6, column 3.
3. The 5 region at the bottom needs a total of exactly 5 (3 of its 5 cells are already filled) — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–3 domino vertically on row 3, column 2 and row 4, column 2.
4. Row 5, column 2 has only one open neighbor, so its domino must run to row 5, column 1; the 5 region at the bottom needs a total of exactly 5 (4 of its 5 cells are already filled) — nothing else fits. Place the 1–5 domino horizontally on row 5, column 1 and row 5, column 2.
5. Row 1, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the “≠” region at the top needs every pip to be different (1 of its 3 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 5–6 domino vertically on row 1, column 2 and row 2, column 2.

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

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

• the “=” region at the top-left: 5, 5, 5, 5, 5
• the 10 region at the top: 4, 6
• the 0 region at the center: 0, 0, 0, 0
• the “=” region at the bottom-right: 1, 1, 1, 1, 1
• the single-cell 3 region at the bottom: 3
• the 4 region at the bottom: 0, 4

1. The 4 region at the bottom needs a total of exactly 4 — nothing else fits. Place the 0–3 domino vertically on row 6, column 3 and row 7, column 3.
2. The 0 region at the center needs a total of exactly 0 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–0 domino horizontally on row 4, column 3 and row 4, column 4.
3. The “=” region at the top-left needs every pip to match — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–5 domino vertically on row 3, column 2 and row 4, column 2.
4. Row 4, column 5 has only one open neighbor, so its domino must run to row 5, column 5; the 0 region at the center needs a total of exactly 0 (3 of its 4 cells are already filled) — nothing else fits. Place the 0–1 domino vertically on row 4, column 5 and row 5, column 5.
5. The “=” region at the top-left needs every pip to match (1 of its 5 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–5 domino vertically on row 1, column 2 and row 2, column 2.
6. The 10 region at the top needs a total of exactly 10 — nothing else fits. Place the 5–6 domino horizontally on row 2, column 3 and row 2, column 4.
7. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the “=” region at the top-left needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 2–5 domino vertically on row 2, column 1 and row 3, column 1.
8. The “=” region at the bottom-right needs every pip to match (1 of its 5 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–1 domino horizontally on row 6, column 4 and row 6, column 5.
9. Row 7, column 4 has only one open neighbor, so its domino must run to row 7, column 5; the 4 region at the bottom needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–4 domino horizontally on row 7, column 4 and row 7, column 5.
10. Row 1, column 4 has only one open neighbor, so its domino must run to row 1, column 3; the 10 region at the top needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–4 domino horizontally on row 1, column 3 and row 1, column 4.
11. Row 6, column 6 has only one open neighbor, so its domino must run to row 5, column 6; the “=” region at the bottom-right needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 1–6 domino vertically on row 5, column 6 and row 6, column 6.