Pips answers for Sunday, August 24, 2025

Start with the “=” region at the bottom-right: all 3 cells must show the same number, and only 4s appear on enough dominoes to fill it.

• the single-cell 6 region at the left: 6
• the “=” region at the center: 1, 1
• the “=” region at the bottom-right: 4, 4, 4

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

Start with the 16 region at the top: the only way to make it with these dominoes is 4 + 4 + 4 + 4.

• the single-cell 1 region at the top: 1
• the 16 region at the top: 4, 4, 4, 4
• the single-cell “>2” region at the top-right: 3
• the 0 region at the left: 0, 0
• the single-cell 0 region at the bottom-left: 0
• the “=” region at the bottom: 2, 2

1. The 16 region at the top needs a total of exactly 16 — 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 1, column 2 and row 1, column 3.
2. The 0 region at the left 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 1 and row 4, column 1.
3. The 16 region at the top needs a total of exactly 16 (1 of its 4 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–4 domino horizontally on row 4, column 2 and row 4, column 3.
4. Row 6, column 1 has only one open neighbor, so its domino must run to row 6, column 2; the single-cell 0 region at the bottom-left needs a total of exactly 0 — nothing else fits. Place the 0–2 domino horizontally on row 6, column 1 and row 6, column 2.
5. Row 6, column 3 has only one open neighbor, so its domino must run to row 5, column 3; the “=” region at the bottom needs every pip to match (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 2–3 domino vertically on row 5, column 3 and row 6, column 3.
6. The 16 region at the top needs a total of exactly 16 (2 of its 4 cells are already filled) — 3 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 2, column 3 and row 2, column 4.
7. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the 16 region at the top needs a total of exactly 16 (3 of its 4 cells are already filled) — nothing else fits. Place the 2–4 domino horizontally on row 3, column 3 and row 3, column 4.

Start with the “=” region at the center: all 7 cells must show the same number, and only 6s appear on enough dominoes to fill it.

• the single-cell “<3” region at the top-left: 2
• the “=” region at the center: 6, 6, 6, 6, 6, 6, 6
• the single-cell “<2” region at the top-right: 0
• the 2 region at the top-right: 1, 1
• the “=” region at the bottom-left: 4, 4
• the 0 region at the bottom-right: 0, 0
• the 10 region at the bottom: 5, 5

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

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