Pips answers for Thursday, April 2, 2026

Start with the “=” region at the right: all 4 cells must show the same number, and only 3s appear on enough dominoes to fill it.

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

1. The “=” region at the 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 3–3 domino vertically on row 2, column 3 and row 3, column 3.
2. The 10 region at the bottom-left needs a total of exactly 10 — nothing else fits. Place the 3–5 domino horizontally on row 4, column 2 and row 4, column 3.
3. The “≠” region at the top-left needs every pip to be different — nothing else fits. Place the 2–3 domino horizontally on row 1, column 2 and row 1, column 3.
4. The “≠” region at the top-left needs every pip to be different (1 of its 4 cells is already filled) — nothing else fits. Place the 0–5 domino vertically on row 3, column 1 and row 4, column 1.
5. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the “≠” region at the top-left needs every pip to be different (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 1–6 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 right: the only way to make it with these dominoes is 6 + 6.

• the 16 region at the top-left: 5, 5, 6
• the 5 region at the top: 1, 2, 2
• the single-cell 4 region at the top-right: 4
• the 12 region at the right: 6, 6
• the 9 region at the bottom-left: 4, 5
• the 6 region at the bottom: 3, 3
• the 1 region at the bottom: 0, 1
• the single-cell 3 region at the bottom-right: 3

1. The 12 region at the right needs a total of exactly 12 — nothing else fits. Place the 6–6 domino vertically on row 2, column 4 and row 3, column 4.
2. The 1 region at the bottom needs a total of exactly 1 — nothing else fits. Place the 1–3 domino horizontally on row 4, column 3 and row 4, column 4.
3. The 5 region at the top needs a total of exactly 5 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–4 domino horizontally on row 1, column 3 and row 1, column 4.
4. The 5 region at the top needs a total of exactly 5 (1 of its 3 cells is already filled) — nothing else fits. Place the 0–2 domino vertically on row 2, column 3 and row 3, column 3.
5. The 5 region at the top needs a total of exactly 5 (2 of its 3 cells are already filled) — nothing else fits. Place the 2–3 domino vertically on row 2, column 2 and row 3, column 2.
6. The 9 region at the bottom-left needs a total of exactly 9 — nothing else fits. Place the 3–5 domino horizontally on row 4, column 1 and row 4, column 2.
7. The 16 region at the top-left needs a total of exactly 16 — nothing else fits. Place the 4–5 domino vertically on row 2, column 1 and row 3, column 1.
8. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the 16 region at the top-left needs a total of exactly 16 (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 5–6 domino horizontally on row 1, column 1 and row 1, column 2.

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

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

• the “=” region at the top: 6, 6
• the single-cell 4 region at the top-right: 4
• the single-cell “<3” region at the top-left: 2
• the “>8” region at the top-left: 4, 5
• the “>9” region at the top-right: 4, 6
• the single-cell 1 region at the center: 1
• the “=” region at the left: 0, 0, 0, 0
• the “≠” region at the center: 1, 2, 3, 4, 5, 6
• the single-cell 5 region at the right: 5
• the “=” region at the right: 2, 2
• the 0 region at the bottom: 0, 0
• the single-cell 1 region at the right: 1
• the 3 region at the bottom-left: 1, 1, 1
• the 7 region at the bottom: 3, 4

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