Pips answers for Tuesday, September 30, 2025

Start with the “≠” region at the center — it has the fewest pip combinations on the board, so the dominoes around it are the easiest to pin down.

• the single-cell “>3” region at the top: 4
• the “≠” region at the center: 0, 2, 3, 4
• the single-cell “<3” region at the right: 2
• the single-cell “>4” region at the bottom: 5

1. The single-cell “>4” region at the bottom must total more than 4 — nothing else fits. Place the 3–5 domino horizontally on row 3, column 2 and row 3, column 3.
2. The single-cell “>3” region at the top must total more than 3 — nothing else fits. Place the 4–4 domino horizontally on row 1, column 2 and row 1, column 3.
3. The “≠” region at the center needs every pip to be different (2 of its 4 cells are already filled) — nothing else fits. Place the 2–2 domino horizontally on row 2, column 3 and row 2, column 4.
4. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, column 1; the “≠” region at the center needs every pip to be different (3 of its 4 cells are already filled) — nothing else fits. Place the 0–3 domino horizontally on row 2, column 1 and row 2, column 2.

Start with the 10 region at the bottom: only 4 + 6, or 5 + 5 can make it with these dominoes.

• the 8 region at the center: 0, 1, 3, 4
• the single-cell “<3” region at the left: 0
• the single-cell “<5” region at the right: 1
• the 10 region at the bottom: 5, 5

1. The 10 region at the bottom needs a total of exactly 10 — 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 vertically on row 4, column 2 and row 5, column 2.
2. Row 6, column 2 has only one open neighbor, so its domino must run to row 6, column 3; the 10 region at the bottom needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–5 domino horizontally on row 6, column 2 and row 6, column 3.
3. The 8 region at the center needs a total of exactly 8 (1 of its 4 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 0–1 domino horizontally on row 3, column 1 and row 3, column 2.
4. Row 3, column 3 has only one open neighbor, so its domino must run to row 4, column 3; the 8 region at the center needs a total of exactly 8 (2 of its 4 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 1–3 domino vertically on row 3, column 3 and row 4, column 3.
5. Row 2, column 2 has only one open neighbor, so its domino must run to row 1, column 2; the 8 region at the center needs a total of exactly 8 (3 of its 4 cells are already filled) — nothing else fits. Place the 4–6 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 bottom-left: the only way to make it with these dominoes is 0 + 0.

• the 10 region at the top-left: 5, 5
• the “=” region at the top: 0, 0, 0, 0, 0, 0
• the “>10” region at the top-right: 6, 6
• the “=” region at the right: 1, 1, 1, 1, 1
• the “=” region at the bottom-left: 3, 3
• the 0 region at the bottom-left: 0, 0
• the “=” region at the bottom-right: 2, 2, 2, 2

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

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