Pips answers for Sunday, December 21, 2025

Start with the 8 region at the bottom-right: the only way to make it with these dominoes is 3 + 5.

• the “=” region at the left: 0, 0, 0
• the single-cell 4 region at the top: 4
• the 6 region at the bottom: 2, 2, 2
• the 8 region at the bottom-right: 3, 5

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

Start with the “=” region at the top-left: all 3 cells must show the same number, and only 2 or 5 or 6 appear on enough dominoes to fill it.

• the “=” region at the top-left: 2, 2, 2
• the “=” region at the top: 5, 5
• the 6 region at the bottom-left: 1, 5
• the 11 region at the right: 1, 4, 6
• the single-cell 6 region at the bottom-left: 6
• the single-cell 6 region at the bottom: 6
• the single-cell “<4” region at the bottom-right: 0

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

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

Start with the “=” region at the bottom-left: all 6 cells must show the same number, and only 1s appear on enough dominoes to fill it.

• the “≠” region at the top-right: 3, 4, 5, 6
• the single-cell 3 region at the top: 3
• the single-cell 2 region at the top-right: 2
• the 11 region at the left: 5, 6
• the 18 region at the center: 6, 6, 6
• the “=” region at the center: 4, 4
• the “=” region at the right: 0, 0, 0, 0, 0
• the single-cell 5 region at the left: 5
• the “=” region at the bottom-left: 1, 1, 1, 1, 1, 1
• the single-cell 5 region at the center: 5

1. The “≠” region at the top-right needs every pip to be different — nothing else fits. Place the 2–4 domino horizontally on row 2, column 6 and row 2, column 7.
2. The 18 region at the center needs a total of exactly 18 — nothing else fits. Place the 5–6 domino horizontally on row 3, column 2 and row 3, column 3.
3. 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 horizontally on row 5, column 6 and row 5, column 7.
4. The single-cell 3 region at the top needs a total of exactly 3 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–6 domino horizontally on row 2, column 4 and row 2, column 5.
5. The “=” region at the right needs every pip to match (2 of its 5 cells are already filled) — 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 6 and row 4, column 6.
6. The 18 region at the center needs a total of exactly 18 (1 of its 3 cells is already filled) — nothing else fits. Place the 0–6 domino vertically on row 3, column 5 and row 4, column 5.
7. The “=” region at the bottom-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 1–6 domino vertically on row 4, column 2 and row 5, column 2.
8. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 18 region at the center needs a total of exactly 18 (2 of its 3 cells are already filled) — nothing else fits. Place the 4–6 domino vertically on row 3, column 4 and row 4, column 4.
9. Row 5, column 1 has only one open neighbor, so its domino must run to row 6, column 1; the single-cell 5 region at the left needs a total of exactly 5 — nothing else fits. Place the 1–5 domino vertically on row 5, column 1 and row 6, column 1.
10. The single-cell 5 region at the center needs a total of exactly 5 — nothing else fits. Place the 4–5 domino horizontally on row 5, column 3 and row 5, column 4.
11. Row 5, column 5 has only one open neighbor, so its domino must run to row 6, column 5; the “=” region at the right needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 0–1 domino vertically on row 5, column 5 and row 6, column 5.
12. The “=” region at the bottom-left needs every pip to match (3 of its 6 cells are already filled) — nothing else fits. Place the 1–1 domino horizontally on row 6, column 3 and row 6, column 4.
13. Row 6, column 2 has only one open neighbor, so its domino must run to row 7, column 2; the “=” region at the bottom-left needs every pip to match (5 of its 6 cells are already filled) — nothing else fits. Place the 1–4 domino vertically on row 6, column 2 and row 7, column 2.
14. Row 1, column 6 has only one open neighbor, so its domino must run to row 1, column 5; the “≠” region at the top-right needs every pip to be different (3 of its 4 cells are already filled) — nothing else fits. Place the 3–4 domino horizontally on row 1, column 5 and row 1, column 6.