Pips answers for Thursday, February 5, 2026

Start with the “=” region at the top: all 3 cells must show the same number, and only 5s appear on enough dominoes to fill it.

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

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

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

Start with the 10 region at the right: the only way to make it with these dominoes is 4 + 6.

• the single-cell 2 region at the top: 2
• the 7 region at the top: 2, 5
• the single-cell 4 region at the left: 4
• the single-cell 3 region at the left: 3
• the single-cell 2 region at the right: 2
• the “=” region at the left: 2, 2
• the 10 region at the right: 4, 6
• the 8 region at the bottom: 2, 6
• the single-cell “>1” region at the bottom: 6

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

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

Start with the 18 region at the bottom-left: the only way to make it with these dominoes is 6 + 6 + 6.

• the “=” region at the top: 4, 4, 4
• the single-cell 5 region at the top-right: 5
• the “=” region at the top-right: 1, 1
• the single-cell 5 region at the left: 5
• the 5 region at the left: 1, 4
• the “=” region at the right: 2, 2
• the 18 region at the bottom-left: 6, 6, 6
• the “=” region at the bottom-left: 0, 0
• the “=” region at the bottom-right: 3, 3, 3, 3
• the 10 region at the bottom-right: 5, 5
• the “=” region at the bottom-right: 0, 0

1. The 18 region at the bottom-left needs a total of exactly 18 — nothing else fits. Place the 6–6 domino vertically on row 3, column 1 and row 4, column 1.
2. Row 4, column 2 has only one open neighbor, so its domino must run to row 4, column 3; the 18 region at the bottom-left needs a total of exactly 18 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–6 domino horizontally on row 4, column 2 and row 4, column 3.
3. The 5 region at the left needs a total of exactly 5 — 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 3 and row 3, column 3.
4. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the single-cell 5 region at the left needs a total of exactly 5 — nothing else fits. Place the 1–5 domino horizontally on row 2, column 1 and row 2, column 2.
5. 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 4–5 domino horizontally on row 1, column 5 and row 1, column 6.
6. The “=” region at the top needs every pip to match (1 of its 3 cells is 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 vertically on row 2, column 5 and row 3, column 5.
7. 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 horizontally on row 2, column 6 and row 2, column 7.
8. The “=” region at the right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 2–5 domino vertically on row 2, column 8 and row 3, column 8.
9. The “=” region at the bottom-right needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 3–3 domino horizontally on row 3, column 6 and row 3, column 7.
10. Row 1, column 7 has only one open neighbor, so its domino must run to row 1, column 8; the “=” region at the top-right needs every pip to match — nothing else fits. Place the 1–1 domino horizontally on row 1, column 7 and row 1, column 8.
11. Row 4, column 5 has only one open neighbor, so its domino must run to row 4, column 6; the “=” region at the bottom-right needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 0–3 domino horizontally on row 4, column 5 and row 4, column 6.
12. Row 4, column 7 has only one open neighbor, so its domino must run to row 4, column 8; the “=” region at the bottom-right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 0–5 domino horizontally on row 4, column 7 and row 4, column 8.