Pips answers for Sunday, October 26, 2025

Start with the 17 region at the center: the only way to make it with these dominoes is 5 + 6 + 6.

• the “>7” region at the top-left: 2, 6
• the “=” region at the top-right: 5, 5
• the 17 region at the center: 5, 6, 6

1. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the “=” region at the top-right 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 5–5 domino vertically on row 1, column 4 and row 2, column 4.
2. The 17 region at the center needs a total of exactly 17 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–5 domino vertically on row 1, column 1 and row 2, column 1.
3. Row 1, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the “>7” region at the top-left must total more than 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 6–6 domino vertically on row 1, column 2 and row 2, column 2.
4. Row 2, column 3 has only one open neighbor, so its domino must run to row 3, column 3; the 17 region at the center needs a total of exactly 17 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–6 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 22 region at the bottom: the only way to make it with these dominoes is 5 + 5 + 6 + 6.

• the “≠” region at the top: 2, 4, 5
• the 0 region at the center: 0, 0
• the single-cell 1 region at the center: 1
• the single-cell 2 region at the center: 2
• the “=” region at the bottom-left: 0, 0, 0
• the single-cell 3 region at the left: 3
• the “=” region at the center: 1, 1, 1, 1
• the 10 region at the bottom-left: 5, 5
• the 22 region at the bottom: 5, 5, 6, 6
• the “≠” region at the bottom-right: 2, 4

1. The single-cell 3 region at the left needs a total of exactly 3 — nothing else fits. Place the 1–3 domino horizontally on row 5, column 3 and row 5, column 4.
2. The 0 region at the center needs a total of exactly 0 — nothing else fits. Place the 0–2 domino vertically on row 3, column 4 and row 4, column 4.
3. The “=” region at the center needs every pip to match (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 1–1 domino horizontally on row 5, column 5 and row 5, column 6.
4. The “=” region at the center needs every pip to match (3 of its 4 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 1–4 domino horizontally on row 5, column 7 and row 5, column 8.
5. The “≠” region at the top needs every pip to be different — nothing else fits. Place the 1–2 domino vertically on row 2, column 5 and row 3, column 5.
6. Row 5, column 1 has only one open neighbor, so its domino must run to row 5, column 2; the “=” region at the bottom-left 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 0–0 domino horizontally on row 5, column 1 and row 5, column 2.
7. The “≠” region at the top needs every pip to be different (1 of its 3 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 0–4 domino vertically on row 2, column 3 and row 3, column 3.
8. Row 6, column 2 has only one open neighbor, so its domino must run to row 6, column 3; the “=” region at the bottom-left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 0–5 domino horizontally on row 6, column 2 and row 6, column 3.
9. The 22 region at the bottom needs a total of exactly 22 — 3 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 7, column 3 and row 7, column 4.
10. Row 6, column 4 has only one open neighbor, so its domino must run to row 6, column 5; the 22 region at the bottom needs a total of exactly 22 (1 of its 4 cells is already filled) — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–5 domino horizontally on row 6, column 4 and row 6, column 5.
11. Row 7, column 5 has only one open neighbor, so its domino must run to row 7, column 6; the 22 region at the bottom needs a total of exactly 22 (3 of its 4 cells are already filled) — nothing else fits. Place the 4–6 domino horizontally on row 7, column 5 and row 7, column 6.
12. Row 2, column 4 has only one open neighbor, so its domino must run to row 1, column 4; the “≠” region at the top needs every pip to be different (2 of its 3 cells are already filled) — nothing else fits. Place the 4–5 domino vertically on row 1, column 4 and row 2, column 4.
13. Row 6, column 6 has only one open neighbor, so its domino must run to row 6, column 7; the “≠” region at the bottom-right needs every pip to be different (1 of its 2 cells is already filled) — nothing else fits. Place the 2–4 domino horizontally on row 6, column 6 and row 6, column 7.

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

Start with the 1 region at the center: the only way to make it with these dominoes is 0 + 1.

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

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