Pips answers for Monday, January 19, 2026

Start with the 7 region at the top-right: the only way to make it with these dominoes is 3 + 4.

• the 3 region at the left: 1, 2
• the 7 region at the top-right: 3, 4
• the 4 region at the bottom: 1, 3

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

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

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

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

Start with the 12 region at the top-right: the only way to make it with these dominoes is 6 + 6.

• the single-cell 1 region at the top: 1
• the “=” region at the top: 2, 2, 2, 2
• the 12 region at the top-right: 6, 6
• the 4 region at the center: 1, 1, 1, 1
• the 7 region at the right: 3, 4
• the “=” region at the left: 0, 0, 0, 0
• the single-cell 2 region at the bottom: 2
• the 10 region at the bottom: 5, 5
• the single-cell “>4” region at the bottom: 6

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

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