Pips answers for Thursday, September 17, 2026

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

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

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

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

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

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

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

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

• the “=” region at the top-left: 5, 5, 5
• the “=” region at the top-right: 2, 2
• the single-cell “>1” region at the top-right: 5
• the 5 region at the left: 1, 1, 1, 1, 1
• the 9 region at the top: 3, 6
• the 0 region at the top-right: 0, 0
• the single-cell “>3” region at the top-right: 4
• the single-cell 4 region at the left: 4
• the single-cell 0 region at the center: 0
• the “≠” region at the bottom-right: 0, 1, 2, 3, 4, 6
• the 11 region at the bottom-left: 5, 6
• the 11 region at the bottom: 5, 6
• the single-cell 3 region at the bottom-left: 3
• the single-cell 4 region at the bottom-left: 4

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

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