Pips answers for Saturday, September 6, 2025

Start with the 10 region at the bottom-left: the only way to make it with these dominoes is 5 + 5.

• the 6 region at the top: 3, 3
• the single-cell 0 region at the top-right: 0
• the 10 region at the bottom-left: 5, 5
• the single-cell 0 region at the center: 0
• the 9 region at the bottom: 3, 6

1. The 10 region at the bottom-left needs a total of exactly 10 — nothing else fits. Place the 3–5 domino horizontally on row 4, column 1 and row 4, column 2.
2. Row 3, column 1 has only one open neighbor, so its domino must run to row 3, column 2; the 10 region at the bottom-left needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–5 domino horizontally on row 3, column 1 and row 3, column 2.
3. Row 4, column 3 has only one open neighbor, so its domino must run to row 3, column 3; the 9 region at the bottom needs a total of exactly 9 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–6 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 — nothing else fits. Place the 3–3 domino vertically on row 1, column 2 and row 2, column 2.
5. Row 1, column 4 has only one open neighbor, so its domino must run to row 1, column 3; the single-cell 0 region at the top-right needs a total of exactly 0 — nothing else fits. Place the 0–2 domino horizontally on row 1, column 3 and row 1, column 4.

Start with the 9 region at the top: the only way to make it with these dominoes is 4 + 5.

• the “=” region at the top-left: 3, 3
• the 9 region at the top: 4, 5
• the “=” region at the bottom-right: 2, 2
• the single-cell “<3” region at the bottom-left: 1
• the single-cell 4 region at the bottom: 4

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

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

• the “=” region at the top: 3, 3, 3
• the 11 region at the top-right: 5, 6
• the “≠” region at the top-right: 0, 1, 2
• the 15 region at the left: 5, 5, 5
• the single-cell 2 region at the left: 2
• the single-cell 4 region at the center: 4
• the single-cell 4 region at the center: 4
• the 12 region at the bottom: 6, 6
• the 1 region at the bottom: 0, 1
• the single-cell 3 region at the bottom: 3

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