Pips answers for Wednesday, November 5, 2025

Start with the 12 region at the top: the only way to make it with these dominoes is 4 + 4 + 4.

• the 12 region at the top: 4, 4, 4
• the 2 region at the bottom: 1, 1
• the 0 region at the right: 0, 0

1. The 2 region at the bottom needs a total of exactly 2 — nothing else fits. Place the 1–1 domino vertically on row 2, column 2 and row 3, column 2.
2. The 0 region at the right needs a total of exactly 0 — nothing else fits. Place the 0–0 domino horizontally on row 2, column 3 and row 2, column 4.
3. The 12 region at the top needs a total of exactly 12 — 3 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 1 and row 2, column 1.
4. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the 12 region at the top needs a total of exactly 12 (1 of its 3 cells is already filled) — nothing else fits. Place the 4–4 domino horizontally on row 1, column 2 and row 1, column 3.

Start with the 12 region at the left: the only way to make it with these dominoes is 6 + 6.

• the 12 region at the left: 6, 6
• the 5 region at the center: 1, 2, 2
• the single-cell “>3” region at the bottom-left: 4
• the single-cell “>3” region at the bottom-right: 5

1. The 12 region at the left needs a total of exactly 12 — 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 1 and row 2, column 1.
2. The 12 region at the left needs a total of exactly 12 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–6 domino horizontally on row 2, column 2 and row 2, column 3.
3. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 3; the single-cell “>3” region at the bottom-left 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 0–4 domino horizontally on row 3, column 2 and row 3, column 3.
4. The single-cell “>3” region at the bottom-right must total more than 3 — nothing else fits. Place the 0–5 domino horizontally on row 3, column 4 and row 3, column 5.
5. The 5 region at the center needs a total of exactly 5 (1 of its 3 cells is already filled) — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–2 domino vertically on row 1, column 4 and row 2, column 4.
6. Row 2, column 5 has only one open neighbor, so its domino must run to row 2, column 6; the 5 region at the center needs a total of exactly 5 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–1 domino horizontally on row 2, column 5 and row 2, column 6.

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

Start with the 10 region at the center: only 4 + 6, or 5 + 5 can make it with these dominoes.

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

1. The single-cell 0 region at the right needs a total of exactly 0 — 3 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 1, column 5 and row 2, column 5.
2. The 10 region at the center needs a total of exactly 10 — 7 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–5 domino vertically on row 2, column 2 and row 3, column 2.
3. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; 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 vertically on row 3, column 1 and row 4, column 1.
4. Row 2, column 3 has only one open neighbor, so its domino must run to row 2, column 4; the 10 region at the center needs a total of exactly 10 (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 5–6 domino horizontally on row 2, column 3 and row 2, column 4.
5. The 10 region at the right needs a total of exactly 10 — 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 3, column 4 and row 3, column 5.
6. The 10 region at the bottom needs a total of exactly 10 — 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 4, column 2 and row 4, column 3.
7. Row 4, column 4 has only one open neighbor, so its domino must run to row 4, column 5; the 10 region at the bottom needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–6 domino horizontally on row 4, column 4 and row 4, column 5.

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