Pips answers for Thursday, September 10, 2026

Start with the “>17” region at the center: the only way to make it with these dominoes is 6 + 6 + 6.

• the “>17” region at the center: 6, 6, 6
• the “=” region at the top-right: 4, 4
• the single-cell “>4” region at the left: 5
• the “=” region at the bottom-right: 0, 0
• the single-cell 3 region at the bottom: 3

1. The “>17” region at the center must total more than 17 — nothing else fits. Place the 5–6 domino horizontally on row 2, column 1 and row 2, column 2.
2. The single-cell 3 region at the bottom needs a total of exactly 3 — nothing else fits. Place the 0–3 domino horizontally on row 3, column 3 and row 3, column 4.
3. The “=” region at the top-right needs every pip to match — nothing else fits. Place the 0–4 domino vertically on row 1, column 4 and row 2, column 4.
4. The “=” region at the top-right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino vertically on row 1, column 3 and row 2, column 3.
5. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 1; the “>17” region at the center must total more than 17 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–6 domino horizontally on row 1, column 1 and row 1, column 2.

Start with the 4 region at the top: only 0 + 4, or 1 + 3, or 2 + 2 can make it with these dominoes.

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

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

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

• the single-cell 0 region at the top-left: 0
• the 15 region at the top-left: 4, 5, 6
• the single-cell “<3” region at the top-right: 2
• the single-cell “>3” region at the top-right: 6
• the single-cell 0 region at the top: 0
• the single-cell “>2” region at the top: 3
• the single-cell 0 region at the top-right: 0
• the single-cell “>4” region at the top-right: 5
• the “<4” region at the right: 0, 0
• the 2 region at the bottom-left: 1, 1
• the 0 region at the bottom-left: 0, 0
• the single-cell “<3” region at the bottom: 1
• the single-cell 3 region at the bottom: 3
• the “=” region at the bottom-right: 5, 5, 5
• the “=” region at the bottom-left: 2, 2, 2, 2
• the 8 region at the bottom: 4, 4
• the single-cell 1 region at the bottom-right: 1

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

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