Pips answers for Friday, January 16, 2026

Start with the “=” region at the top-left: all 3 cells must show the same number, and only 0 or 2 appear on enough dominoes to fill it.

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

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

Start with the 7 region at the center: the only way to make it with these dominoes is 2 + 2 + 3.

• the “≠” region at the top: 3, 4, 5, 6
• the “<11” region at the left: 5, 5
• the 7 region at the center: 2, 2, 3
• the single-cell “>4” region at the bottom: 6

1. The 7 region at the center needs a total of exactly 7 — 6 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–6 domino vertically on row 5, column 3 and row 6, column 3.
2. The “≠” region at the top needs every pip to be different — 6 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 3, column 3 and row 4, column 3.
3. Row 4, column 4 has only one open neighbor, so its domino must run to row 4, column 5; the 7 region at the center needs a total of exactly 7 (2 of its 3 cells are already filled) — nothing else fits. Place the 2–4 domino horizontally on row 4, column 4 and row 4, column 5.
4. Row 4, column 1 has only one open neighbor, so its domino must run to row 4, column 2; the “<11” region at the left must total less than 11 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–5 domino horizontally on row 4, column 1 and row 4, column 2.
5. The “≠” region at the top needs every pip to be different (1 of its 4 cells is already filled) — 5 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–6 domino horizontally on row 2, column 1 and row 2, column 2.
6. The “≠” region at the top needs every pip to be different (2 of its 4 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 6–6 domino vertically on row 1, column 3 and row 2, column 3.
7. Row 2, column 4 has only one open neighbor, so its domino must run to row 2, column 5; the “≠” region at the top needs every pip to be different (3 of its 4 cells are already filled) — nothing else fits. Place the 5–6 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 24 region at the top: the only way to make it with these dominoes is 6 + 6 + 6 + 6.

• the 10 region at the top: 3, 3, 4
• the single-cell 1 region at the top-right: 1
• the 24 region at the top: 6, 6, 6, 6
• the “=” region at the center: 4, 4, 4, 4, 4
• the single-cell 1 region at the left: 1
• the “≠” region at the right: 0, 1, 2, 3, 4, 5
• the 20 region at the bottom-left: 5, 5, 5, 5
• the 0 region at the bottom-left: 0, 0
• the 5 region at the bottom: 2, 3
• the 5 region at the bottom-left: 2, 3

1. The 24 region at the top needs a total of exactly 24 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino vertically on row 1, column 6 and row 2, column 6.
2. The 24 region at the top needs a total of exactly 24 (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 6–6 domino horizontally on row 2, column 4 and row 2, column 5.
3. The 10 region at the top needs a total of exactly 10 — nothing else fits. Place the 4–6 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 1, column 5; the 10 region at the top needs a total of exactly 10 (1 of its 3 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 3–3 domino horizontally on row 1, column 4 and row 1, column 5.
5. The 0 region at the bottom-left needs a total of exactly 0 — 3 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 7, column 1 and row 7, column 2.
6. Row 8, column 1 has only one open neighbor, so its domino must run to row 8, column 2; the 0 region at the bottom-left needs a total of exactly 0 (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 0–3 domino horizontally on row 8, column 1 and row 8, column 2.
7. The “=” region at the center 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 4–4 domino horizontally on row 3, column 3 and row 3, column 4.
8. The “=” region at the center needs every pip to match (2 of its 5 cells are already filled) — nothing else fits. Place the 1–4 domino horizontally on row 4, column 3 and row 4, column 4.
9. The “=” region at the center needs every pip to match (3 of its 5 cells are already filled) — nothing else fits. Place the 4–5 domino horizontally on row 5, column 3 and row 5, column 4.
10. The 20 region at the bottom-left needs a total of exactly 20 (2 of its 4 cells are already filled) — nothing else fits. Place the 5–5 domino vertically on row 6, column 3 and row 7, column 3.
11. Row 6, column 4 has only one open neighbor, so its domino must run to row 7, column 4; the “=” region at the center needs every pip to match (4 of its 5 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 2–4 domino vertically on row 6, column 4 and row 7, column 4.
12. Row 8, column 3 has only one open neighbor, so its domino must run to row 8, column 4; 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 2–3 domino horizontally on row 8, column 3 and row 8, column 4.
13. The “≠” region at the right needs every pip to be different — 6 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 4, column 7 and row 4, column 8.
14. Row 5, column 7 has only one open neighbor, so its domino must run to row 6, column 7; the “≠” region at the right needs every pip to be different (2 of its 6 cells are already filled) — 4 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 5, column 7 and row 6, column 7.
15. Row 7, column 6 has only one open neighbor, so its domino must run to row 7, column 7; the “≠” region at the right needs every pip to be different (4 of its 6 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 0–4 domino horizontally on row 7, column 6 and row 7, column 7.

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