Pips answers for Sunday, September 7, 2025

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

• the 0 region at the left: 0, 0, 0
• the single-cell 1 region at the center: 1
• the 2 region at the center: 1, 1
• the 3 region at the right: 1, 2

1. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the 0 region at the left needs a total of exactly 0 — nothing else fits. Place the 0–0 domino horizontally on row 1, column 1 and row 1, column 2.
2. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the 0 region at the left needs a total of exactly 0 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–1 domino horizontally on row 1, column 3 and row 1, column 4.
3. Row 1, column 5 has only one open neighbor, so its domino must run to row 1, column 6; the 2 region at the center needs a total of exactly 2 — nothing else fits. Place the 1–1 domino horizontally on row 1, column 5 and row 1, column 6.
4. Row 1, column 7 has only one open neighbor, so its domino must run to row 1, column 8; the 3 region at the right needs a total of exactly 3 — 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 1, column 7 and row 1, column 8.

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

Start with the “=” region at the bottom: all 4 cells must show the same number, and only 3s appear on enough dominoes to fill it.

• the 10 region at the top: 4, 6
• the 11 region at the top-right: 5, 6
• the 10 region at the left: 5, 5
• the “=” region at the bottom: 3, 3, 3, 3

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

Start with the “>15” region at the bottom: the only way to make it with these dominoes is 5 + 5 + 6.

• the 10 region at the top-left: 5, 5
• the “=” region at the top-right: 4, 4, 4
• the 2 region at the top-left: 1, 1
• the 6 region at the left: 2, 4
• the “≠” region at the center: 2, 3, 4, 5
• the “=” region at the right: 3, 3, 3
• the “<3” region at the right: 0, 0, 2
• the “>15” region at the bottom: 5, 5, 6
• the 0 region at the bottom-right: 0, 0
• the single-cell 1 region at the bottom: 1

1. Row 7, column 3 has only one open neighbor, so its domino must run to row 7, column 4; the “>15” region at the bottom must total more than 15 — nothing else fits. Place the 5–5 domino horizontally on row 7, column 3 and row 7, column 4.
2. Row 1, column 9 has only one open neighbor, so its domino must run to row 2, column 9; the “=” region at the top-right needs every pip to match — nothing else fits. Place the 4–4 domino vertically on row 1, column 9 and row 2, column 9.
3. The “≠” region at the center needs every pip to be different — nothing else fits. Place the 2–6 domino vertically on row 5, column 4 and row 6, column 4.
4. Row 7, column 6 has only one open neighbor, so its domino must run to row 7, column 7; the single-cell 1 region at the bottom needs a total of exactly 1 — nothing else fits. Place the 0–1 domino horizontally on row 7, column 6 and row 7, column 7.
5. The “<3” region at the right must total less than 3 — nothing else fits. Place the 0–2 domino vertically on row 5, column 7 and row 6, column 7.
6. The “=” region at the right 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 0–3 domino vertically on row 3, column 7 and row 4, column 7.
7. Row 3, column 8 has only one open neighbor, so its domino must run to row 3, column 9; the “=” region at the right needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 3–4 domino horizontally on row 3, column 8 and row 3, column 9.
8. The “=” region at the right needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 3–5 domino horizontally on row 3, column 5 and row 3, column 6.
9. The “≠” region at the center needs every pip to be different (2 of its 4 cells are already filled) — nothing else fits. Place the 0–4 domino horizontally on row 4, column 5 and row 4, column 6.
10. Row 4, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the “≠” region at the center needs every pip to be different (3 of its 4 cells are already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–3 domino vertically on row 3, column 4 and row 4, column 4.
11. The 2 region at the top-left needs a total of exactly 2 — nothing else fits. Place the 1–4 domino vertically on row 2, column 3 and row 3, column 3.
12. The 10 region at the top-left needs a total of exactly 10 — nothing else fits. Place the 1–5 domino vertically on row 1, column 2 and row 2, column 2.
13. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 10 region at the top-left needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–5 domino vertically on row 1, column 1 and row 2, column 1.