Pips answers for Monday, December 22, 2025

Start with the 7 region at the center: only 1 + 6, or 2 + 5 can make it with these dominoes.

• the 6 region at the top: 0, 6
• the 7 region at the center: 1, 6
• the single-cell 4 region at the right: 4
• the single-cell 1 region at the left: 1
• the single-cell “>1” region at the bottom: 6

1. The single-cell 4 region at the right needs a total of exactly 4 — nothing else fits. Place the 4–5 domino horizontally on row 2, column 4 and row 2, column 5.
2. Row 3, column 1 has only one open neighbor, so its domino must run to row 3, column 2; the single-cell 1 region at the left needs a total of exactly 1 — 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 3, column 1 and row 3, column 2.
3. The 7 region at the center needs a total of exactly 7 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino vertically on row 3, column 3 and row 4, column 3.
4. 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 0–6 domino vertically on row 1, column 3 and row 2, column 3.
5. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 1; 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 2–6 domino horizontally on row 1, column 1 and row 1, column 2.

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

Start with the 9 region at the right: only 3 + 6, or 4 + 5 can make it with these dominoes.

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

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

Start with the 30 region at the bottom: the only way to make it with these dominoes is 6 + 6 + 6 + 6 + 6.

• the 0 region at the top-left: 0, 0, 0, 0
• the single-cell “<4” region at the top-left: 1
• the single-cell 4 region at the top: 4
• the single-cell 0 region at the top: 0
• the 12 region at the top-right: 6, 6
• the single-cell 0 region at the top-right: 0
• the “=” region at the center: 2, 2, 2, 2
• the “>8” region at the top: 5, 5
• the single-cell “<4” region at the top-right: 3
• the 4 region at the right: 1, 1, 1, 1
• the 30 region at the bottom: 6, 6, 6, 6, 6
• the 4 region at the bottom-right: 2, 2
• the “=” region at the bottom-left: 3, 3, 3
• the single-cell “>4” region at the bottom-right: 5

1. The 0 region at the top-left needs a total of exactly 0 — 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 2, column 1 and row 3, column 1.
2. The single-cell 4 region at the top needs a total of exactly 4 — nothing else fits. Place the 2–4 domino vertically on row 1, column 3 and row 2, column 3.
3. Row 4, column 1 has only one open neighbor, so its domino must run to row 4, column 2; the 0 region at the top-left needs a total of exactly 0 (2 of its 4 cells are already filled) — nothing else fits. Place the 0–2 domino horizontally on row 4, column 1 and row 4, column 2.
4. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the 0 region at the top-left needs a total of exactly 0 (3 of its 4 cells are already filled) — nothing else fits. Place the 0–1 domino horizontally on row 1, column 1 and row 1, column 2.
5. The single-cell 0 region at the top-right needs a total of exactly 0 — nothing else fits. Place the 0–6 domino horizontally on row 1, column 5 and row 1, column 6.
6. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the single-cell 0 region at the top needs a total of exactly 0 — nothing else fits. Place the 0–5 domino vertically on row 1, column 4 and row 2, column 4.
7. The 30 region at the bottom needs a total of exactly 30 — nothing else fits. Place the 6–6 domino horizontally on row 5, column 2 and row 5, column 3.
8. The “=” region at the center 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 2–2 domino vertically on row 3, column 3 and row 4, column 3.
9. Row 6, column 1 has only one open neighbor, so its domino must run to row 6, column 2; 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 3–3 domino horizontally on row 6, column 1 and row 6, column 2.
10. Row 6, column 3 has only one open neighbor, so its domino must run to row 6, column 4; the “=” region at the bottom-left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 3–6 domino horizontally on row 6, column 3 and row 6, column 4.
11. The 30 region at the bottom needs a total of exactly 30 (3 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 2–6 domino horizontally on row 5, column 4 and row 5, column 5.
12. Row 6, column 5 has only one open neighbor, so its domino must run to row 6, column 6; the 4 region at the bottom-right needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–5 domino horizontally on row 6, column 5 and row 6, column 6.
13. The 30 region at the bottom needs a total of exactly 30 (4 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 5–6 domino vertically on row 3, column 4 and row 4, column 4.
14. The 12 region at the top-right needs a total of exactly 12 (1 of its 2 cells is already filled) — 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 2, column 5 and row 3, column 5.
15. Row 4, column 5 has only one open neighbor, so its domino must run to row 4, column 6; the 4 region at the right needs a total of exactly 4 (1 of its 4 cells is already filled) — nothing else fits. Place the 1–1 domino horizontally on row 4, column 5 and row 4, column 6.
16. Row 2, column 6 has only one open neighbor, so its domino must run to row 3, column 6; the single-cell “<4” region at the top-right must total less than 4 — nothing else fits. Place the 1–3 domino vertically on row 2, column 6 and row 3, column 6.