Pips answers for Thursday, December 11, 2025

Start with the 9 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 5 region at the top-left: 5
• the “>3” region at the top-right: 2, 2
• the 9 region at the center: 1, 2, 6
• the single-cell “>3” region at the bottom: 4

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

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

Start with the 2 region at the top-left: the only way to make it with these dominoes is 1 + 1.

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

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

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

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

• the “=” region at the top: 2, 2, 2
• the single-cell “>1” region at the top-right: 5
• the “=” region at the top-left: 1, 1
• the 10 region at the center: 5, 5
• the 9 region at the right: 3, 6
• the 9 region at the bottom-right: 2, 3, 4
• the 10 region at the bottom-left: 4, 6
• the 0 region at the bottom: 0, 0
• the “=” region at the bottom: 1, 1, 1, 1

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