Pips answers for Thursday, September 25, 2025

Start with the “=” region at the bottom: all 2 cells must show the same number, and only 1 or 3 appear on enough dominoes to fill it.

• the 10 region at the left: 3, 3, 4
• the single-cell “>2” region at the right: 3
• the single-cell 5 region at the bottom-right: 5
• the “=” region at the bottom: 1, 1

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

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

• the “=” region at the top: 0, 0, 0
• the “=” region at the center: 3, 3, 3, 3, 3
• the 3 region at the right: 1, 2
• the 1 region at the left: 0, 1
• the 11 region at the right: 5, 6
• the single-cell 1 region at the bottom-left: 1
• the 10 region at the bottom: 5, 5
• the “=” region at the bottom-right: 2, 2

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

Start with the 10 region at the top-right: the only way to make it with these dominoes is 5 + 5.

• the single-cell “>1” region at the top-left: 2
• the “=” region at the top: 3, 3, 3, 3, 3
• the 10 region at the top-right: 5, 5
• the 8 region at the bottom-left: 2, 6
• the 2 region at the bottom: 1, 1
• the 7 region at the bottom-right: 1, 6
• the 6 region at the bottom-right: 0, 6

1. The 10 region at the top-right needs a total of exactly 10 — 4 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 1, column 3 and row 1, column 4.
2. The “=” region at the top needs every pip to match — nothing else fits. Place the 3–5 domino horizontally on row 2, column 3 and row 2, column 4.
3. The “=” region at the top needs every pip to match (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 2–3 domino horizontally on row 1, column 1 and row 1, column 2.
4. The “=” region at the top needs every pip to match (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 1–3 domino vertically on row 3, column 2 and row 4, column 2.
5. The “=” region at the top needs every pip to match (3 of its 5 cells are already filled) — nothing else fits. Place the 3–6 domino vertically on row 3, column 1 and row 4, column 1.
6. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, column 1; the “=” region at the top needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 0–3 domino horizontally on row 2, column 1 and row 2, column 2.
7. The 8 region at the bottom-left needs a total of exactly 8 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–2 domino vertically on row 5, column 1 and row 6, column 1.
8. The 2 region at the bottom needs a total of exactly 2 (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 1–1 domino horizontally on row 5, column 2 and row 5, column 3.
9. Row 5, column 4 has only one open neighbor, so its domino must run to row 6, column 4; the 7 region at the bottom-right needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 6–6 domino vertically on row 5, column 4 and row 6, column 4.
10. Row 6, column 3 has only one open neighbor, so its domino must run to row 6, column 2; the 6 region at the bottom-right needs a total of exactly 6 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–0 domino horizontally on row 6, column 2 and row 6, column 3.