Pips answers for Tuesday, September 29, 2026

Start with the “=” region at the center: all 2 cells must show the same number, and only 0 or 2 or 3 or 5 appear on enough dominoes to fill it.

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

1. 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-right must total less than 2 — nothing else fits. Place the 0–5 domino vertically on row 1, column 3 and row 2, column 3.
2. The “=” region at the center needs every pip to match — nothing else fits. Place the 2–5 domino horizontally on row 4, column 2 and row 4, column 3.
3. Row 2, column 2 has only one open neighbor, so its domino must run to row 3, column 2; the single-cell “<2” region at the top must total less than 2 — nothing else fits. Place the 0–2 domino vertically on row 2, column 2 and row 3, column 2.
4. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; the single-cell “<2” region at the left must total less than 2 — nothing else fits. Place the 0–3 domino vertically on row 3, column 1 and row 4, column 1.
5. Row 5, column 1 has only one open neighbor, so its domino must run to row 5, 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 3–4 domino horizontally on row 5, column 1 and row 5, column 2.

Start with the “=” region at the top-left: all 3 cells must show the same number, and only 4 or 6 appear on enough dominoes to fill it.

• the “=” region at the top-left: 4, 4, 4
• the “<7” region at the top: 0, 5
• the “≠” region at the center: 2, 6
• the “<12” region at the right: 3, 3, 5
• the “>8” region at the right: 6, 6
• the single-cell 1 region at the left: 1

1. The single-cell 1 region at the left needs a total of exactly 1 — 2 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 3, column 1 and row 3, column 2.
2. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; 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 4–4 domino horizontally on row 2, column 1 and row 2, column 2.
3. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the “=” region at the top-left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 4–5 domino horizontally on row 1, column 2 and row 1, column 3.
4. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the “<7” region at the top must total less than 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–5 domino vertically on row 1, column 4 and row 2, column 4.
5. Row 2, column 5 has only one open neighbor, so its domino must run to row 3, column 5; the “>8” region at the right must total more than 8 — nothing else fits. Place the 6–6 domino vertically on row 2, column 5 and row 3, column 5.
6. The “<12” region at the right must total less than 12 (1 of its 3 cells is already filled) — nothing else fits. Place the 3–3 domino vertically on row 3, column 4 and row 4, column 4.
7. 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 be different — 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 2, column 3 and row 3, column 3.

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

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

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

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

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