Pips answers for Sunday, September 21, 2025

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

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

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

Start with the 42 region at the bottom: the only way to make it with these dominoes is 6 + 6 + 6 + 6 + 6 + 6 + 6.

• the 4 region at the top-left: 2, 2
• the “=” region at the top-right: 5, 5
• the single-cell “<6” region at the top-left: 5
• the single-cell 0 region at the top-right: 0
• the 42 region at the bottom: 6, 6, 6, 6, 6, 6, 6
• the single-cell “<2” region at the center: 1
• the single-cell “<4” region at the bottom-left: 3
• the single-cell “<5” region at the bottom: 4
• the single-cell “<3” region at the bottom-right: 2

1. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 4 region at the top-left needs a total of exactly 4 — nothing else fits. Place the 2–2 domino vertically on row 1, column 1 and row 2, column 1.
2. The single-cell 0 region at the top-right needs a total of exactly 0 — nothing else fits. Place the 0–6 domino vertically on row 2, column 5 and row 3, column 5.
3. The 42 region at the bottom needs a total of exactly 42 (1 of its 7 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 3 and row 3, column 4.
4. Row 1, column 6 has only one open neighbor, so its domino must run to row 2, column 6; 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 5–5 domino vertically on row 1, column 6 and row 2, column 6.
5. The single-cell “<4” region at the bottom-left must total less than 4 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–6 domino vertically on row 4, column 2 and row 5, column 2.
6. The single-cell “<3” region at the bottom-right must total less than 3 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–6 domino vertically on row 4, column 5 and row 5, column 5.
7. Row 3, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the 42 region at the bottom needs a total of exactly 42 (3 of its 7 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–6 domino vertically on row 2, column 2 and row 3, column 2.
8. The 42 region at the bottom needs a total of exactly 42 (4 of its 7 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–6 domino vertically on row 4, column 3 and row 5, column 3.
9. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the 42 region at the bottom needs a total of exactly 42 (5 of its 7 cells are already filled) — nothing else fits. Place the 6–6 domino vertically on row 4, column 4 and row 5, column 4.

Start with the 18 region at the right: the only way to make it with these dominoes is 6 + 6 + 6.

• the “=” region at the top: 4, 4
• the single-cell 3 region at the top-left: 3
• the single-cell 5 region at the top: 5
• the single-cell 1 region at the top: 1
• the 18 region at the right: 6, 6, 6
• the single-cell 3 region at the left: 3
• the single-cell “<4” region at the left: 1
• the “≠” region at the center: 2, 3, 4, 5, 6
• the 10 region at the right: 5, 5
• the single-cell “>4” region at the left: 6
• the single-cell 2 region at the bottom-left: 2
• the 0 region at the bottom: 0, 0, 0
• the single-cell 4 region at the bottom: 4
• the single-cell 1 region at the bottom-right: 1

1. The single-cell 3 region at the top-left needs a total of exactly 3 — nothing else fits. Place the 1–3 domino vertically on row 2, column 2 and row 3, column 2.
2. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; the single-cell 3 region at the left needs a total of exactly 3 — nothing else fits. Place the 3–6 domino vertically on row 3, column 1 and row 4, column 1.
3. The single-cell 1 region at the bottom-right needs a total of exactly 1 — nothing else fits. Place the 1–6 domino vertically on row 4, column 5 and row 5, column 5.
4. The 18 region at the right needs a total of exactly 18 (1 of its 3 cells is already filled) — nothing else fits. Place the 6–6 domino vertically on row 2, column 5 and row 3, column 5.
5. Row 3, column 6 has only one open neighbor, so its domino must run to row 4, column 6; the 10 region at the right needs a total of exactly 10 — nothing else fits. Place the 5–5 domino vertically on row 3, column 6 and row 4, column 6.
6. The 0 region at the bottom needs a total of exactly 0 — nothing else fits. Place the 0–0 domino horizontally on row 6, column 3 and row 6, column 4.
7. The 0 region at the bottom needs a total of exactly 0 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–6 domino vertically on row 4, column 3 and row 5, column 3.
8. Row 4, column 2 has only one open neighbor, so its domino must run to row 5, column 2; the “≠” region at the center needs every pip to be different (1 of its 5 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–5 domino vertically on row 4, column 2 and row 5, column 2.
9. The “≠” region at the center needs every pip to be different (2 of its 5 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–4 domino vertically on row 4, column 4 and row 5, column 4.
10. The “=” region at the top 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 4–4 domino horizontally on row 1, column 3 and row 1, column 4.
11. The single-cell 5 region at the top needs a total of exactly 5 — nothing else fits. Place the 4–5 domino vertically on row 2, column 3 and row 3, column 3.
12. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the single-cell 1 region at the top needs a total of exactly 1 — nothing else fits. Place the 1–2 domino vertically on row 2, column 4 and row 3, column 4.