Pips answers for Monday, January 12, 2026

Start with the “=” region at the top: all 2 cells must show the same number, and only 4 or 5 appear on enough dominoes to fill it.

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

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

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

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

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

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

Start with the 2 region at the bottom-left: the only way to make it with these dominoes is 0 + 0 + 1 + 1.

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

1. The 12 region at the top-right needs a total of exactly 12 — 4 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 6 and row 2, column 6.
2. The 10 region at the top-right needs a total of exactly 10 — nothing else fits. Place the 5–6 domino vertically on row 1, column 5 and row 2, column 5.
3. The 4 region at the top needs a total of exactly 4 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–5 domino vertically on row 1, column 4 and row 2, column 4.
4. The 10 region at the left needs a total of exactly 10 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–5 domino vertically on row 5, column 1 and row 6, column 1.
5. The 10 region at the left 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.
6. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the 4 region at the top 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 1, column 3 and row 2, column 3.
7. The “=” region at the center needs every pip to match (2 of its 4 cells are already filled) — nothing else fits. Place the 3–6 domino horizontally on row 4, column 3 and row 4, column 4.
8. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the “=” region at the center needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 3–4 domino horizontally on row 3, column 3 and row 3, column 4.
9. Row 5, column 4 has only one open neighbor, so its domino must run to row 6, column 4; the 12 region at the center needs a total of exactly 12 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–6 domino vertically on row 5, column 4 and row 6, column 4.
10. The 4 region at the bottom needs a total of exactly 4 (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 0–2 domino horizontally on row 7, column 3 and row 7, column 4.
11. Row 6, column 2 has only one open neighbor, so its domino must run to row 6, column 3; the 2 region at the bottom-left needs a total of exactly 2 (2 of its 4 cells are already filled) — nothing else fits. Place the 1–1 domino horizontally on row 6, column 2 and row 6, column 3.
12. Row 8, column 3 has only one open neighbor, so its domino must run to row 8, column 4 — nothing else fits. Place the 4–4 domino horizontally on row 8, column 3 and row 8, column 4.