Pips answers for Thursday, September 11, 2025

Start with the “=” region at the top: all 3 cells must show the same number, and only 1s appear on enough dominoes to fill it.

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

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

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

• the single-cell 3 region at the top-left: 3
• the 14 region at the top: 4, 5, 5
• the 15 region at the top-right: 3, 6, 6
• the 9 region at the left: 4, 5
• the single-cell 2 region at the right: 2
• the single-cell 6 region at the bottom-left: 6
• the single-cell 5 region at the bottom-right: 5

1. The 15 region at the top-right needs a total of exactly 15 — nothing else fits. Place the 2–3 domino vertically on row 2, column 5 and row 3, column 5.
2. The 15 region at the top-right needs a total of exactly 15 (1 of its 3 cells is already filled) — nothing else fits. Place the 6–6 domino horizontally on row 1, column 5 and row 1, column 6.
3. The 9 region at the left needs a total of exactly 9 — nothing else fits. Place the 5–6 domino vertically on row 4, column 2 and row 5, column 2.
4. The 14 region at the top needs a total of exactly 14 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–4 domino horizontally on row 1, column 1 and row 1, column 2.
5. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the 14 region at the top needs a total of exactly 14 (1 of its 3 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 1, column 3 and row 1, column 4.
6. Row 3, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the 9 region at the left needs a total of exactly 9 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–4 domino vertically on row 2, column 2 and row 3, column 2.
7. Row 5, column 5 has only one open neighbor, so its domino must run to row 4, column 5; the single-cell 5 region at the bottom-right needs a total of exactly 5 — nothing else fits. Place the 3–5 domino vertically on row 4, column 5 and row 5, column 5.

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

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

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