Pips answers for Monday, August 25, 2025

Start with the 15 region at the left: the only way to make it with these dominoes is 5 + 5 + 5.

• the single-cell 2 region at the top-left: 2
• the 15 region at the left: 5, 5, 5
• the 3 region at the right: 1, 1, 1

1. The 15 region at the left needs a total of exactly 15 — nothing else fits. Place the 2–5 domino vertically on row 1, column 1 and row 2, column 1.
2. The 3 region at the right needs a total of exactly 3 — nothing else fits. Place the 1–1 domino vertically on row 2, column 4 and row 3, column 4.
3. 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 1–5 domino horizontally on row 2, column 2 and row 2, column 3.
4. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the 15 region at the left needs a total of exactly 15 (2 of its 3 cells are already filled) — nothing else fits. Place the 3–5 domino horizontally on row 1, column 2 and row 1, column 3.

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

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

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

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

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

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