Pips answers for Saturday, October 10, 2026

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

• the single-cell 1 region at the top: 1
• the 10 region at the center: 4, 6
• the “=” region at the left: 0, 0
• the single-cell “>2” region at the center: 4
• the 10 region at the bottom: 5, 5
• the single-cell “>1” region at the bottom-right: 6

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

Start with the 9 region at the bottom: only 3 + 6, or 4 + 5 can make it with these dominoes.

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

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

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

Start with the single-cell 3 region at the top-left: the only way to make it with these dominoes is 3.

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

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