Pips answers for Friday, December 5, 2025

Start with the 6 region at the bottom: the only way to make it with these dominoes is 2 + 4.

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

1. Row 3, column 1 has only one open neighbor, so its domino must run to row 3, column 2; the single-cell 2 region at the bottom-left needs a total of exactly 2 — nothing else fits. Place the 1–2 domino horizontally on row 3, column 1 and row 3, column 2.
2. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, column 3; the “<2” region at the bottom must total less than 2 (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 0–4 domino horizontally on row 2, column 2 and row 2, column 3.
3. 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-right needs a total of exactly 0 — nothing else fits. Place the 0–0 domino vertically on row 1, column 4 and row 2, column 4.
4. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the 6 region at the bottom needs a total of exactly 6 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–3 domino horizontally on row 3, column 3 and row 3, column 4.

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

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

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

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

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

• the “=” region at the left: 3, 3, 3, 3, 3
• the “=” region at the top-left: 4, 4, 4
• the “=” region at the top-right: 1, 1
• the “≠” region at the top-right: 3, 5
• the single-cell 0 region at the top: 0
• the “=” region at the center: 2, 2
• the 24 region at the right: 6, 6, 6, 6
• the single-cell “<5” region at the bottom-left: 1

1. The 24 region at the right needs a total of exactly 24 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 6–6 domino horizontally on row 3, column 7 and row 3, column 8.
2. The 24 region at the right needs a total of exactly 24 (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 1–6 domino vertically on row 2, column 6 and row 3, column 6.
3. Row 4, column 6 has only one open neighbor, so its domino must run to row 5, column 6; the 24 region at the right needs a total of exactly 24 (3 of its 4 cells are already filled) — nothing else fits. Place the 4–6 domino vertically on row 4, column 6 and row 5, column 6.
4. Row 1, column 6 has only one open neighbor, so its domino must run to row 1, column 7; the “=” region at the top-right needs every pip to match (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 1–5 domino horizontally on row 1, column 6 and row 1, column 7.
5. The single-cell 0 region at the top 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 0–2 domino vertically on row 2, column 4 and row 3, column 4.
6. The “=” region at the left needs every pip to match — nothing else fits. Place the 2–3 domino horizontally on row 3, column 2 and row 3, column 3.
7. The “=” region at the left needs every pip to match (1 of its 5 cells is already filled) — nothing else fits. Place the 3–3 domino vertically on row 2, column 1 and row 3, column 1.
8. The “=” region at the top-left needs every pip to match — nothing else fits. Place the 4–4 domino horizontally on row 1, column 3 and row 1, column 4.
9. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the “=” region at the left needs every pip to match (3 of its 5 cells are already filled) — nothing else fits. Place the 3–4 domino horizontally on row 1, column 1 and row 1, column 2.
10. Row 4, column 1 has only one open neighbor, so its domino must run to row 5, column 1; the “=” region at the left needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 1–3 domino vertically on row 4, column 1 and row 5, column 1.
11. Row 1, column 8 has only one open neighbor, so its domino must run to row 2, column 8; the “≠” region at the top-right needs every pip to be different (1 of its 2 cells is already filled) — nothing else fits. Place the 3–5 domino vertically on row 1, column 8 and row 2, column 8.