Pips answers for Wednesday, January 14, 2026

Start with the 10 region at the bottom-left: only 4 + 6, or 5 + 5 can make it with these dominoes.

• the “=” region at the top-right: 5, 5
• the 10 region at the bottom-left: 4, 6
• the “=” region at the center: 4, 4
• the single-cell 0 region at the bottom: 0

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

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

• the single-cell 2 region at the top: 2
• the “=” region at the top-right: 2, 2, 2
• the 13 region at the left: 1, 3, 3, 6
• the “>9” region at the top: 5, 5
• the single-cell 1 region at the right: 1
• the single-cell “>3” region at the bottom-right: 4
• the single-cell 2 region at the bottom-left: 2

1. The “>9” region at the top must total more than 9 — nothing else fits. Place the 1–5 domino horizontally on row 3, column 3 and row 3, column 4.
2. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the single-cell 2 region at the top needs a total of exactly 2 — nothing else fits. Place the 2–5 domino vertically on row 1, column 3 and row 2, column 3.
3. 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 — 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 5 and row 2, column 5.
4. The single-cell “>3” region at the bottom-right must total more than 3 — 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 5, column 5 and row 6, column 5.
5. The 13 region at the left needs a total of exactly 13 — 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 vertically on row 5, column 1 and row 6, column 1.
6. Row 3, column 5 has only one open neighbor, so its domino must run to row 4, column 5; the “=” region at the top-right needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 0–2 domino vertically on row 3, column 5 and row 4, column 5.
7. The 13 region at the left needs a total of exactly 13 (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 1–3 domino vertically on row 3, column 1 and row 4, column 1.
8. Row 2, column 1 has only one open neighbor, so its domino must run to row 1, column 1; the 13 region at the left needs a total of exactly 13 (3 of its 4 cells are already filled) — nothing else fits. Place the 6–6 domino vertically on row 1, column 1 and row 2, column 1.

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

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

• the 12 region at the top-left: 4, 4, 4
• the “≠” region at the top-right: 0, 1
• the single-cell 2 region at the top-right: 2
• the “=” region at the top-left: 1, 1
• the 12 region at the left: 6, 6
• the single-cell “>2” region at the right: 3
• the 12 region at the right: 6, 6
• the 0 region at the left: 0, 0
• the 15 region at the center: 5, 5, 5
• the 2 region at the bottom-left: 1, 1
• the single-cell 1 region at the right: 1
• the “=” region at the bottom-left: 3, 3, 3
• the “=” region at the bottom-right: 3, 3
• the single-cell “>3” region at the bottom-right: 4
• the single-cell “<3” region at the bottom-right: 1

1. The 12 region at the right needs a total of exactly 12 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 6–6 domino vertically on row 3, column 5 and row 4, column 5.
2. The 12 region at the left needs a total of exactly 12 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–6 domino vertically on row 3, column 1 and row 4, column 1.
3. The 12 region at the left needs a total of exactly 12 (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 vertically on row 2, column 2 and row 3, column 2.
4. The 0 region at the left needs a total of exactly 0 (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 0–3 domino vertically on row 5, column 1 and row 6, column 1.
5. The 2 region at the bottom-left needs a total of exactly 2 — nothing else fits. Place the 1–1 domino vertically on row 5, column 2 and row 6, column 2.
6. Row 7, column 1 has only one open neighbor, so its domino must run to row 7, column 2; the “=” region at the bottom-left needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 3–3 domino horizontally on row 7, column 1 and row 7, column 2.
7. The 12 region at the top-left 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 1–4 domino vertically on row 1, column 1 and row 2, column 1.
8. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the 12 region at the top-left needs a total of exactly 12 (1 of its 3 cells is already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–4 domino horizontally on row 1, column 2 and row 1, column 3.
9. The 15 region at the center needs a total of exactly 15 — 4 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–5 domino vertically on row 3, column 4 and row 4, column 4.
10. Row 4, column 3 has only one open neighbor, so its domino must run to row 4, column 2; the 15 region at the center needs a total of exactly 15 (1 of its 3 cells is already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–5 domino horizontally on row 4, column 2 and row 4, column 3.
11. The 15 region at the center needs a total of exactly 15 (2 of its 3 cells are 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 4 and row 5, column 5.
12. The single-cell “>3” region at the bottom-right must total more than 3 — nothing else fits. Place the 3–4 domino vertically on row 6, column 4 and row 7, column 4.
13. Row 6, column 5 has only one open neighbor, so its domino must run to row 7, column 5; the “=” region at the bottom-right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 1–3 domino vertically on row 6, column 5 and row 7, column 5.
14. The single-cell 2 region at the top-right needs a total of exactly 2 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–2 domino horizontally on row 1, column 4 and row 1, column 5.
15. Row 2, column 4 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 be different (1 of its 2 cells is already filled) — nothing else fits. Place the 0–0 domino horizontally on row 2, column 4 and row 2, column 5.