Pips answers for Friday, February 6, 2026

Start with the 10 region at the bottom: the only way to make it with these dominoes is 5 + 5.

• the “=” region at the top-right: 2, 2, 2
• the single-cell 5 region at the top-right: 5
• the single-cell “<1” region at the center: 0
• the “=” region at the bottom-left: 6, 6
• the 10 region at the bottom: 5, 5
• the single-cell “>5” region at the bottom-right: 6

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

Start with the 4 region at the left: the only way to make it with these dominoes is 1 + 1 + 2.

• the single-cell “>3” region at the top-left: 4
• the “=” region at the top-right: 5, 5
• the 4 region at the left: 1, 1, 2
• the 16 region at the center: 5, 5, 6
• the single-cell 1 region at the right: 1
• the 21 region at the bottom: 4, 5, 6, 6

1. The 4 region at the left needs a total of exactly 4 — 3 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 1, column 1 and row 2, column 1.
2. The single-cell 1 region at the right needs a total of exactly 1 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–3 domino vertically on row 2, column 5 and row 3, column 5.
3. The 4 region at the left needs a total of exactly 4 (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 2–2 domino vertically on row 3, column 1 and row 4, column 1.
4. 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.
5. The 4 region at the left needs a total of exactly 4 (2 of its 3 cells are already filled) — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–5 domino vertically on row 2, column 2 and row 3, column 2.
6. Row 2, column 3 has only one open neighbor, so its domino must run to row 3, column 3; the 16 region at the center needs a total of exactly 16 (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 5–6 domino vertically on row 2, column 3 and row 3, column 3.
7. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 21 region at the bottom needs a total of exactly 21 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–5 domino vertically on row 3, column 4 and row 4, column 4.
8. Row 4, column 2 has only one open neighbor, so its domino must run to row 4, column 3; the 21 region at the bottom needs a total of exactly 21 (2 of its 4 cells are already filled) — nothing else fits. Place the 6–6 domino horizontally on row 4, column 2 and row 4, column 3.

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

Start with the 10 region at the top: only 4 + 6, or 5 + 5 can make it with these dominoes.

• the 8 region at the top-left: 4, 4
• the 10 region at the top: 5, 5
• the single-cell 1 region at the top-right: 1
• the “=” region at the left: 0, 0
• the 4 region at the left: 2, 2
• the single-cell 4 region at the top: 4
• the 10 region at the top-right: 4, 6
• the “=” region at the bottom: 3, 3, 3, 3
• the “>9” region at the right: 6, 6
• the 10 region at the bottom-left: 5, 5
• the “=” region at the bottom-left: 1, 1
• the “=” region at the bottom-right: 0, 0

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