Pips answers for Monday, August 3, 2026

Start with the 11 region at the left: the only way to make it with these dominoes is 5 + 6.

• the “=” region at the center: 4, 4, 4
• the 11 region at the left: 5, 6
• the single-cell “<3” region at the center: 2
• the single-cell “>3” region at the right: 6
• the 10 region at the bottom-left: 5, 5
• the single-cell “<2” region at the bottom: 0

1. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the “=” region at the center needs every pip to match — nothing else fits. Place the 4–4 domino vertically on row 1, column 3 and row 2, column 3.
2. The “=” region at the center needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 4–6 domino horizontally on row 3, column 3 and row 3, column 4.
3. The 10 region at the bottom-left needs a total of exactly 10 — nothing else fits. Place the 0–5 domino horizontally on row 4, column 2 and row 4, column 3.
4. The 11 region at the left needs a total of exactly 11 — nothing else fits. Place the 2–5 domino horizontally on row 2, column 1 and row 2, column 2.
5. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; the 11 region at the left needs a total of exactly 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 5–6 domino vertically on row 3, column 1 and row 4, column 1.

Start with the 8 region at the center: only 2 + 6, or 3 + 5 can make it with these dominoes.

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

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

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

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

• the 4 region at the top-left: 2, 2
• the 1 region at the top-left: 0, 0, 1
• the 10 region at the top: 5, 5
• the 6 region at the top-right: 2, 4
• the “=” region at the top-right: 3, 3, 3
• the 9 region at the left: 4, 5
• the 12 region at the center: 6, 6
• the single-cell 0 region at the right: 0
• the single-cell 5 region at the right: 5
• the 3 region at the right: 1, 1, 1
• the 7 region at the bottom-left: 3, 4
• the “>10” region at the bottom-left: 6, 6
• the single-cell 0 region at the bottom-right: 0
• the single-cell 1 region at the bottom-left: 1
• the single-cell 1 region at the bottom: 1
• the 10 region at the bottom-right: 4, 6

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