Pips answers for Saturday, October 25, 2025

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

• the “=” region at the top-left: 5, 5, 5
• the “<8” region at the right: 0, 0, 1, 2
• the 11 region at the bottom-left: 5, 6
• the 10 region at the bottom-right: 4, 6

1. Row 1, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the “=” region at the top-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 5–5 domino vertically on row 1, column 2 and row 2, column 2.
2. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the “=” region at the top-left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 5–6 domino vertically on row 2, column 1 and row 3, column 1.
3. Row 4, column 1 has only one open neighbor, so its domino must run to row 4, column 2; the 11 region at the bottom-left needs a total of exactly 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–5 domino horizontally on row 4, column 1 and row 4, column 2.
4. The 10 region at the bottom-right needs a total of exactly 10 — 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 vertically on row 3, column 3 and row 4, column 3.
5. Row 2, column 3 has only one open neighbor, so its domino must run to row 2, column 4; the “<8” region at the right must total less than 8 (1 of its 4 cells is already filled) — nothing else fits. Place the 0–0 domino horizontally on row 2, column 3 and row 2, column 4.
6. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the “<8” region at the right must total less than 8 (3 of its 4 cells are already filled) — nothing else fits. Place the 2–6 domino vertically on row 3, column 4 and row 4, column 4.

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

Start with the 24 region at the left: the only way to make it with these dominoes is 6 + 6 + 6 + 6.

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

1. Row 1, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the 24 region at the left needs a total of exactly 24 — nothing else fits. Place the 6–6 domino vertically on row 1, column 2 and row 2, column 2.
2. The “=” region at the 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 1–5 domino vertically on row 1, column 6 and row 2, column 6.
3. The 24 region at the left needs a total of exactly 24 (2 of its 4 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–6 domino vertically on row 3, column 1 and row 4, column 1.
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 3–6 domino horizontally on row 3, column 2 and row 3, column 3.
5. The “≠” region at the bottom needs every pip to be different (1 of its 5 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–2 domino horizontally on row 4, column 2 and row 4, column 3.
6. The “≠” region at the bottom needs every pip to be different (2 of its 5 cells are already filled) — nothing else fits. Place the 5–5 domino horizontally on row 3, column 4 and row 3, column 5.
7. 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 be different (3 of its 5 cells are already filled) — nothing else fits. Place the 1–1 domino horizontally on row 4, column 4 and row 4, column 5.
8. Row 2, column 4 has only one open neighbor, so its domino must run to row 1, column 4; the “≠” region at the bottom needs every pip to be different (4 of its 5 cells are already filled) — nothing else fits. Place the 4–4 domino vertically on row 1, column 4 and row 2, column 4.
9. The “=” region at the right needs every pip to match (2 of its 4 cells are already filled) — nothing else fits. Place the 2–5 domino vertically on row 3, column 6 and row 4, column 6.
10. Row 3, column 7 has only one open neighbor, so its domino must run to row 4, column 7; the “=” region at the right needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 3–5 domino vertically on row 3, column 7 and row 4, column 7.

Start with the 12 region at the top-right: the only way to make it with these dominoes is 6 + 6.

• the single-cell 2 region at the top-left: 2
• the 26 region at the top: 5, 5, 5, 5, 6
• the 12 region at the top-right: 6, 6
• the “≠” region at the top-left: 0, 3, 4
• the single-cell 1 region at the top-right: 1
• the single-cell 4 region at the left: 4
• the “=” region at the right: 2, 2, 2, 2
• the single-cell “>4” region at the bottom-left: 5
• the 0 region at the bottom: 0, 0
• the 4 region at the bottom-right: 1, 1, 1, 1
• the single-cell 0 region at the bottom-left: 0
• the 6 region at the bottom: 2, 4

1. The single-cell “>4” region at the bottom-left must total more than 4 — 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 4, column 2 and row 5, column 2.
2. The 12 region at the top-right needs a total of exactly 12 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–6 domino horizontally on row 1, column 8 and row 1, column 9.
3. The “≠” region at the top-left needs every pip to be different — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–4 domino vertically on row 2, column 2 and row 3, column 2.
4. The 12 region at the top-right needs a total of exactly 12 (1 of its 2 cells is already filled) — 3 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 1, column 7 and row 2, column 7.
5. The 0 region at the bottom 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 0–6 domino vertically on row 3, column 4 and row 4, column 4.
6. 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 (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 0–2 domino horizontally on row 5, column 4 and row 5, column 5.
7. The 4 region at the bottom-right needs a total of exactly 4 — 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 5, column 6 and row 5, column 7.
8. The “=” region at the right needs every pip to match — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–2 domino horizontally on row 3, column 7 and row 3, column 8.
9. Row 3, column 9 has only one open neighbor, so its domino must run to row 4, column 9; the “=” region at the right needs every pip to match (2 of its 4 cells are already filled) — nothing else fits. Place the 1–2 domino vertically on row 3, column 9 and row 4, column 9.
10. Row 5, column 8 has only one open neighbor, so its domino must run to row 5, column 9; the 4 region at the bottom-right needs a total of exactly 4 (2 of its 4 cells are already filled) — nothing else fits. Place the 1–1 domino horizontally on row 5, column 8 and row 5, column 9.
11. Row 3, column 5 has only one open neighbor, so its domino must run to row 3, column 6; the 26 region at the top needs a total of exactly 26 (1 of its 5 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–5 domino horizontally on row 3, column 5 and row 3, column 6.
12. The 26 region at the top needs a total of exactly 26 (2 of its 5 cells are already filled) — 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 1, column 4 and row 2, column 4.
13. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the single-cell 2 region at the top-left needs a total of exactly 2 — nothing else fits. Place the 2–4 domino vertically on row 1, column 3 and row 2, column 3.
14. Row 1, column 5 has only one open neighbor, so its domino must run to row 1, column 6; the 26 region at the top needs a total of exactly 26 (4 of its 5 cells are already filled) — nothing else fits. Place the 1–5 domino horizontally on row 1, column 5 and row 1, column 6.
15. Row 2, column 1 has only one open neighbor, so its domino must run to row 1, column 1; the “≠” region at the top-left needs every pip to be different (2 of its 3 cells are already filled) — nothing else fits. Place the 0–3 domino vertically on row 1, column 1 and row 2, column 1.