Pips answers for Monday, January 5, 2026

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

• the single-cell 6 region at the top-left: 6
• the “=” region at the top: 2, 2
• the single-cell “>3” region at the top-right: 4
• the 12 region at the right: 6, 6
• the 11 region at the bottom-right: 5, 6

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

Start with the “=” region at the right: all 4 cells must show the same number, and only 0 or 6 appear on enough dominoes to fill it.

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

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

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

Start with the 11 region at the top-right: the only way to make it with these dominoes is 5 + 6.

• the 11 region at the top-left: 3, 4, 4
• the single-cell “<1” region at the top: 0
• the 11 region at the top-right: 5, 6
• the single-cell “>1” region at the top-right: 5
• the “=” region at the right: 2, 2, 2
• the 11 region at the bottom-right: 3, 3, 5
• the 1 region at the bottom-left: 0, 1
• the 11 region at the bottom-left: 5, 6
• the single-cell 1 region at the bottom: 1

1. The 11 region at the top-right needs a total of exactly 11 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–5 domino vertically on row 1, column 6 and row 2, column 6.
2. Row 1, column 7 has only one open neighbor, so its domino must run to row 1, column 8; the 11 region at the top-right needs a total of exactly 11 (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 1–6 domino horizontally on row 1, column 7 and row 1, column 8.
3. The 11 region at the bottom-left needs a total of exactly 11 — nothing else fits. Place the 1–5 domino horizontally on row 5, column 3 and row 5, column 4.
4. The 1 region at the bottom-left needs a total of exactly 1 — nothing else fits. Place the 0–6 domino horizontally on row 5, column 1 and row 5, column 2.
5. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the single-cell “<1” region at the top must total less than 1 — nothing else fits. Place the 0–1 domino vertically on row 1, column 4 and row 2, column 4.
6. The 11 region at the top-left needs a total of exactly 11 — 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.
7. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 11 region at the top-left needs a total of exactly 11 (1 of its 3 cells is already filled) — nothing else fits. Place the 4–4 domino vertically on row 1, column 1 and row 2, column 1.
8. Row 2, column 8 has only one open neighbor, so its domino must run to row 3, column 8; the “=” region at the right needs every pip to match — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–2 domino vertically on row 2, column 8 and row 3, column 8.
9. The 11 region at the bottom-right needs a total of exactly 11 — nothing else fits. Place the 2–5 domino horizontally on row 3, column 6 and row 3, column 7.
10. Row 4, column 6 has only one open neighbor, so its domino must run to row 5, column 6; the 11 region at the bottom-right needs a total of exactly 11 (1 of its 3 cells is already filled) — nothing else fits. Place the 3–3 domino vertically on row 4, column 6 and row 5, column 6.
11. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4 — nothing else fits. Place the 1–1 domino vertically on row 3, column 4 and row 4, column 4.