Pips answers for Tuesday, February 24, 2026

Start with the 6 region at the top-right: only 0 + 2 + 4, or 2 + 2 + 2 can make it with these dominoes.

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

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

Start with the “=” region at the right: all 3 cells must show the same number, and only 1 or 2 appear on enough dominoes to fill it.

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

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

Start with the 4 region at the top-right: only 0 + 4, or 1 + 3, or 2 + 2 can make it with these dominoes.

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

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

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