Pips answers for Thursday, October 23, 2025

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

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

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

Start with the 27 region at the bottom-right: the only way to make it with these dominoes is 3 + 6 + 6 + 6 + 6.

• the 9 region at the top: 1, 3, 5
• the “<2” region at the top-right: 0, 1
• the “=” region at the bottom-left: 2, 2, 2, 2
• the 27 region at the bottom-right: 3, 6, 6, 6, 6
• the single-cell “>4” region at the right: 5

1. The “=” region at the bottom-left needs every pip to match — 6 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–2 domino vertically on row 4, column 2 and row 5, column 2.
2. The “=” region at the bottom-left needs every pip to match (2 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 0–2 domino horizontally on row 3, column 1 and row 3, column 2.
3. The “=” region at the bottom-left 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 2–3 domino horizontally on row 5, column 3 and row 5, column 4.
4. The 27 region at the bottom-right needs a total of exactly 27 (1 of its 5 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 6–6 domino horizontally on row 4, column 4 and row 4, column 5.
5. The 27 region at the bottom-right needs a total of exactly 27 (3 of its 5 cells are already filled) — nothing else fits. Place the 5–6 domino horizontally on row 3, column 4 and row 3, column 5.
6. The 9 region at the top needs a total of exactly 9 — nothing else fits. Place the 3–6 domino vertically on row 3, column 3 and row 4, column 3.
7. The 9 region at the top needs a total of exactly 9 (1 of its 3 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 1–1 domino horizontally on row 1, column 3 and row 1, column 4.
8. The 9 region at the top needs a total of exactly 9 (2 of its 3 cells are already filled) — nothing else fits. Place the 3–5 domino horizontally on row 2, column 2 and row 2, column 3.
9. Row 2, column 4 has only one open neighbor, so its domino must run to row 2, column 5; the “<2” region at the top-right must total less than 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–1 domino horizontally on row 2, column 4 and row 2, column 5.

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

Start with the “=” region at the top-left: all 6 cells must show the same number, and only 3s appear on enough dominoes to fill it.

• the “=” region at the top-left: 3, 3, 3, 3, 3, 3
• the “<7” region at the top: 0, 1, 1, 4
• the single-cell “<2” region at the top-right: 1
• the single-cell 2 region at the top-left: 2
• the 24 region at the top-right: 6, 6, 6, 6
• the 26 region at the center: 5, 5, 5, 5, 6
• the “=” region at the left: 4, 4
• the “=” region at the right: 2, 2, 2
• the 1 region at the bottom: 0, 0, 0, 1

1. The 24 region at the top-right needs a total of exactly 24 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 6–6 domino vertically on row 2, column 8 and row 3, column 8.
2. The 1 region at the bottom needs a total of exactly 1 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–0 domino horizontally on row 6, column 4 and row 6, column 5.
3. The 24 region at the top-right needs a total of exactly 24 (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 4–6 domino horizontally on row 2, column 6 and row 2, column 7.
4. Row 1, column 6 has only one open neighbor, so its domino must run to row 1, column 7; the “<7” region at the top must total less than 7 (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 1, column 6 and row 1, column 7.
5. The “<7” region at the top must total less than 7 (2 of its 4 cells are already filled) — 4 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 2, column 5 and row 3, column 5.
6. The “<7” region at the top must total less than 7 (3 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 0–3 domino horizontally on row 2, column 3 and row 2, column 4.
7. The “=” region at the top-left needs every pip to match (1 of its 6 cells is already filled) — nothing else fits. Place the 3–3 domino horizontally on row 1, column 2 and row 1, column 3.
8. The “=” region at the top-left needs every pip to match (3 of its 6 cells are already filled) — nothing else fits. Place the 2–3 domino horizontally on row 2, column 1 and row 2, column 2.
9. The “=” region at the top-left needs every pip to match (4 of its 6 cells are already filled) — 2 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 3, column 1 and row 3, column 2.
10. The “=” region at the top-left needs every pip to match (5 of its 6 cells are already filled) — nothing else fits. Place the 3–6 domino horizontally on row 3, column 3 and row 3, column 4.
11. The 24 region at the top-right needs a total of exactly 24 (3 of its 4 cells are already filled) — nothing else fits. Place the 2–6 domino vertically on row 3, column 7 and row 4, column 7.
12. Row 3, column 6 has only one open neighbor, so its domino must run to row 4, column 6; the 26 region at the center needs a total of exactly 26 (2 of its 5 cells are already filled) — nothing else fits. Place the 2–5 domino vertically on row 3, column 6 and row 4, column 6.
13. The 26 region at the center needs a total of exactly 26 (3 of its 5 cells are already filled) — nothing else fits. Place the 5–5 domino horizontally on row 4, column 4 and row 4, column 5.
14. Row 4, column 2 has only one open neighbor, so its domino must run to row 4, column 3; the “=” region at the left needs every pip to match — nothing else fits. Place the 4–4 domino horizontally on row 4, column 2 and row 4, column 3.
15. The 1 region at the bottom needs a total of exactly 1 (2 of its 4 cells are already filled) — nothing else fits. Place the 0–2 domino horizontally on row 5, column 5 and row 5, column 6.
16. Row 5, column 4 has only one open neighbor, so its domino must run to row 5, column 3; the 1 region at the bottom needs a total of exactly 1 (3 of its 4 cells are already filled) — nothing else fits. Place the 1–4 domino horizontally on row 5, column 3 and row 5, column 4.