Pips answers for Sunday, August 31, 2025

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

• the “=” region at the top-right: 1, 1
• the “=” region at the bottom-right: 2, 2
• the 10 region at the bottom: 5, 5

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

Start with the “=” region at the bottom-left: all 2 cells must show the same number, and only 1 or 4 appear on enough dominoes to fill it.

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

1. 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 1–6 domino vertically on row 2, column 1 and row 3, column 1.
2. The “>6” region at the center must total more than 6 — nothing else fits. Place the 1–2 domino vertically on row 2, column 2 and row 3, column 2.
3. The “>6” region at the center must total more than 6 (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 4–5 domino horizontally on row 2, column 3 and row 2, column 4.
4. Row 1, column 4 has only one open neighbor, so its domino must run to row 1, column 3; the “>6” region at the top-right must total more than 6 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–4 domino horizontally on row 1, column 3 and row 1, column 4.

Start with the “=” region at the bottom: all 5 cells must show the same number, and only 4s appear on enough dominoes to fill it.

• the 13 region at the top: 2, 2, 2, 3, 4
• the “=” region at the top-right: 1, 1, 1
• the single-cell 0 region at the top-right: 0
• the 24 region at the left: 6, 6, 6, 6
• the 4 region at the center: 1, 3
• the single-cell 5 region at the left: 5
• the “=” region at the bottom: 4, 4, 4, 4, 4
• the single-cell 0 region at the bottom: 0

1. The 24 region at the left needs a total of exactly 24 — nothing else fits. Place the 6–6 domino horizontally on row 3, column 2 and row 3, column 3.
2. Row 4, column 1 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 5–6 domino horizontally on row 4, column 1 and row 4, column 2.
3. Row 2, column 3 has only one open neighbor, so its domino must run to row 2, column 4; the 24 region at the left needs a total of exactly 24 (3 of its 4 cells are already filled) — nothing else fits. Place the 4–6 domino horizontally on row 2, column 3 and row 2, column 4.
4. The “=” region at the bottom needs every pip to match — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–4 domino vertically on row 6, column 3 and row 7, column 3.
5. The “=” region at the top-right needs every pip to match — nothing else fits. Place the 0–1 domino horizontally on row 1, column 6 and row 1, column 7.
6. The “=” region at the top-right needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 1–1 domino horizontally on row 2, column 5 and row 2, column 6.
7. The “=” region at the bottom needs every pip to match (1 of its 5 cells is already filled) — nothing else fits. Place the 4–4 domino vertically on row 5, column 4 and row 6, column 4.
8. The “=” region at the bottom needs every pip to match (3 of its 5 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 2–4 domino horizontally on row 5, column 2 and row 5, column 3.
9. The 4 region at the center needs a total of exactly 4 — nothing else fits. Place the 3–4 domino vertically on row 4, column 5 and row 5, column 5.
10. The 13 region at the top needs a total of exactly 13 (1 of its 5 cells is already filled) — nothing else fits. Place the 1–2 domino horizontally on row 3, column 4 and row 3, column 5.
11. Row 1, column 4 has only one open neighbor, so its domino must run to row 1, column 5; the 13 region at the top needs a total of exactly 13 (2 of its 5 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 3–3 domino horizontally on row 1, column 4 and row 1, column 5.
12. Row 4, column 3 has only one open neighbor, so its domino must run to row 4, column 4; the 13 region at the top needs a total of exactly 13 (3 of its 5 cells are already filled) — nothing else fits. Place the 2–2 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.