Pips answers for Saturday, August 30, 2025

Start with the 16 region at the right: the only way to make it with these dominoes is 5 + 5 + 6.

• the 1 region at the top: 0, 1
• the 16 region at the right: 5, 5, 6
• the 7 region at the left: 3, 4

1. The 1 region at the top needs a total of exactly 1 — nothing else fits. Place the 0–6 domino horizontally on row 1, column 3 and row 1, column 4.
2. Row 1, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the 1 region at the top needs a total of exactly 1 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–3 domino vertically on row 1, column 2 and row 2, column 2.
3. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the 7 region at the left needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–4 domino vertically on row 2, column 1 and row 3, column 1.
4. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the 16 region at the right needs a total of exactly 16 (1 of its 3 cells is already filled) — nothing else fits. Place the 5–5 domino vertically on row 2, column 4 and row 3, column 4.
5. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 3 — nothing else fits. Place the 2–2 domino horizontally on row 3, column 2 and row 3, column 3.

Start with the “<3” region at the bottom — it has the fewest pip combinations on the board, so the dominoes around it are the easiest to pin down.

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

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

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

Start with the “=” region at the center: all 7 cells must show the same number, and only 2s appear on enough dominoes to fill it.

• the 11 region at the top-left: 5, 6
• the single-cell 1 region at the top: 1
• the single-cell 4 region at the top: 4
• the 12 region at the top-right: 6, 6
• the single-cell 4 region at the top: 4
• the “=” region at the center: 2, 2, 2, 2, 2, 2, 2
• the 7 region at the bottom-left: 1, 6
• the single-cell 4 region at the bottom: 4
• the “=” region at the bottom-right: 3, 3, 3
• the 4 region at the bottom-right: 0, 4
• the 0 region at the bottom: 0, 0, 0

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