Pips answers for Friday, January 23, 2026

Start with the 9 region at the left: only 3 + 6, or 4 + 5 can make it with these dominoes.

• the single-cell 5 region at the top-right: 5
• the 9 region at the left: 4, 5
• the “=” region at the center: 3, 3
• the 9 region at the right: 3, 6
• the “=” region at the bottom-left: 0, 0

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

Start with the 1 region at the bottom-left: the only way to make it with these dominoes is 0 + 1.

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

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

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

• the 7 region at the top-right: 1, 1, 5
• the 5 region at the top: 1, 2, 2
• the 7 region at the center: 2, 5
• the 6 region at the center: 3, 3
• the 10 region at the center: 5, 5
• the 9 region at the bottom: 3, 6
• the 3 region at the bottom: 1, 2

1. The 9 region at the bottom needs a total of exactly 9 — 5 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino horizontally on row 7, column 4 and row 7, column 5.
2. The 3 region at the bottom needs a total of exactly 3 (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 2–3 domino horizontally on row 6, column 4 and row 6, column 5.
3. The 6 region at the center needs a total of exactly 6 — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–3 domino vertically on row 4, column 4 and row 5, column 4.
4. The 10 region at the center needs a total of exactly 10 — nothing else fits. Place the 5–5 domino vertically on row 4, column 5 and row 5, column 5.
5. The 7 region at the top-right needs a total of exactly 7 — 8 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 7 and row 1, column 8.
6. Row 2, column 7 has only one open neighbor, so its domino must run to row 2, column 6; the 7 region at the top-right needs a total of exactly 7 (2 of its 3 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 0–5 domino horizontally on row 2, column 6 and row 2, column 7.
7. The 7 region at the center needs a total of exactly 7 — 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 horizontally on row 3, column 5 and row 3, column 6.
8. The 5 region at the top needs a total of exactly 5 — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–1 domino horizontally on row 2, column 2 and row 2, column 3.
9. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the 5 region at the top needs a total of exactly 5 (1 of its 3 cells is already filled) — nothing else fits. Place the 2–2 domino horizontally on row 3, column 3 and row 3, column 4.
10. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2 — nothing else fits. Place the 0–0 domino horizontally on row 1, column 1 and row 1, column 2.

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