Pips answers for Monday, September 22, 2025

Start with the 1 region at the center: the only way to make it with these dominoes is 0 + 1.

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

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

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

• the 8 region at the top-right: 2, 6
• the 11 region at the top-right: 5, 6
• the “=” region at the center: 1, 1, 1
• the 9 region at the bottom-left: 4, 5
• the 9 region at the right: 4, 5

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

Start with the 0 region at the center: the only way to make it with these dominoes is 0 + 0 + 0 + 0 + 0.

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

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