Pips answers for Saturday, December 27, 2025

Start with the 5 region at the top-right: the only way to make it with these dominoes is 0 + 5.

• the “=” region at the top-left: 6, 6
• the 6 region at the top: 2, 4
• the 5 region at the top-right: 0, 5
• the 8 region at the bottom-right: 2, 6
• the 10 region at the bottom-left: 5, 5
• the 4 region at the bottom: 0, 4

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

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

Start with the 4 region at the top-left: only 0 + 4, or 1 + 3, or 2 + 2 can make it with these dominoes.

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

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

Start with the 11 region at the top-left: the only way to make it with these dominoes is 5 + 6.

• the 11 region at the top-left: 5, 6
• the single-cell 2 region at the top-left: 2
• the single-cell “>3” region at the top-left: 5
• the “=” region at the top-left: 3, 3, 3, 3
• the “=” region at the right: 0, 0, 0
• the 12 region at the left: 6, 6
• the single-cell “>2” region at the left: 5
• the “≠” region at the center: 1, 2, 4
• the 0 region at the left: 0, 0
• the single-cell 5 region at the center: 5
• the 10 region at the right: 2, 4, 4
• the single-cell 4 region at the bottom-left: 4
• the single-cell 1 region at the bottom: 1
• the single-cell 5 region at the bottom-left: 5
• the single-cell 2 region at the bottom-left: 2
• the 2 region at the bottom-left: 1, 1

1. The 11 region at the top-left needs a total of exactly 11 — nothing else fits. Place the 5–5 domino vertically on row 1, column 2 and row 2, column 2.
2. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 11 region at the top-left needs a total of exactly 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–6 domino vertically on row 1, column 1 and row 2, column 1.
3. The 12 region at the left needs a total of exactly 12 — nothing else fits. Place the 0–6 domino horizontally on row 5, column 1 and row 5, column 2.
4. The single-cell 4 region at the bottom-left needs a total of exactly 4 — nothing else fits. Place the 0–4 domino horizontally on row 6, column 1 and row 6, column 2.
5. Row 3, column 6 has only one open neighbor, so its domino must run to row 4, column 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 0–0 domino vertically on row 3, column 6 and row 4, column 6.
6. Row 4, column 7 has only one open neighbor, so its domino must run to row 5, column 7; the “=” region at the right needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 0–2 domino vertically on row 4, column 7 and row 5, column 7.
7. The 2 region at the bottom-left needs a total of exactly 2 — nothing else fits. Place the 1–2 domino vertically on row 7, column 2 and row 8, column 2.
8. Row 7, column 1 has only one open neighbor, so its domino must run to row 8, column 1; the single-cell 5 region at the bottom-left needs a total of exactly 5 — nothing else fits. Place the 1–5 domino vertically on row 7, column 1 and row 8, column 1.
9. 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 3–6 domino vertically on row 3, column 1 and row 4, column 1.
10. The “=” region at the top-left needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 3–5 domino vertically on row 3, column 2 and row 4, column 2.
11. Row 5, column 6 has only one open neighbor, so its domino must run to row 6, column 6; the 10 region at the right needs a total of exactly 10 (1 of its 3 cells is already filled) — nothing else fits. Place the 4–4 domino vertically on row 5, column 6 and row 6, column 6.
12. The “=” region at the top-left 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 1–3 domino vertically on row 3, column 3 and row 4, column 3.
13. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the “=” region at the top-left needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 2–3 domino vertically on row 3, column 4 and row 4, column 4.
14. The single-cell 5 region at the center needs a total of exactly 5 — nothing else fits. Place the 2–5 domino horizontally on row 5, column 3 and row 5, column 4.
15. Row 6, column 3 has only one open neighbor, so its domino must run to row 6, column 4; the “≠” region at the center needs every pip to be different (2 of its 3 cells are already filled) — nothing else fits. Place the 1–4 domino horizontally on row 6, column 3 and row 6, column 4.