Pips answers for Wednesday, December 10, 2025

Start with the 30 region at the top: the only way to make it with these dominoes is 5 + 5 + 5 + 5 + 5 + 5.

• the single-cell 1 region at the top-left: 1
• the 30 region at the top: 5, 5, 5, 5, 5, 5
• the single-cell “>3” region at the top-right: 4
• the single-cell “<3” region at the bottom: 2

1. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the single-cell 1 region at the top-left needs a total of exactly 1 — nothing else fits. Place the 1–5 domino horizontally on row 1, column 1 and row 1, column 2.
2. The single-cell “<3” region at the bottom 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 2–5 domino vertically on row 2, column 3 and row 3, column 3.
3. The 30 region at the top needs a total of exactly 30 (2 of its 6 cells are already filled) — 4 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–5 domino horizontally on row 1, column 5 and row 1, column 6.
4. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the 30 region at the top needs a total of exactly 30 (3 of its 6 cells are already filled) — nothing else fits. Place the 5–5 domino horizontally on row 1, column 3 and row 1, column 4.
5. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the 30 region at the top needs a total of exactly 30 (5 of its 6 cells are already filled) — nothing else fits. Place the 3–5 domino vertically on row 2, column 4 and row 3, column 4.

Start with the 30 region at the center: only 4 + 5 + 5 + 5 + 5 + 6, or 5 + 5 + 5 + 5 + 5 + 5 can make it with these dominoes.

• the single-cell 3 region at the top-right: 3
• the single-cell “>3” region at the top-left: 4
• the “=” region at the top-right: 4, 4, 4
• the 30 region at the center: 5, 5, 5, 5, 5, 5
• the single-cell 0 region at the right: 0
• the 7 region at the bottom-right: 1, 6
• the “=” region at the bottom-left: 3, 3
• the “=” region at the bottom: 0, 0

1. Row 2, column 3 has only one open neighbor, so its domino must run to row 2, column 4; 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 4–4 domino horizontally on row 2, column 3 and row 2, column 4.
2. Row 1, column 5 has only one open neighbor, so its domino must run to row 2, column 5; the single-cell 3 region at the top-right needs a total of exactly 3 — nothing else fits. Place the 3–4 domino vertically on row 1, column 5 and row 2, column 5.
3. The single-cell 0 region at the right needs a total of exactly 0 — 6 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 3, column 4 and row 3, column 5.
4. The 30 region at the center needs a total of exactly 30 (1 of its 6 cells is already filled) — 6 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–5 domino vertically on row 2, column 1 and row 3, column 1.
5. The 7 region at the bottom-right needs a total of exactly 7 — 5 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 4 and row 4, column 5.
6. The “=” region at the bottom needs every pip to match — nothing else fits. Place the 0–1 domino horizontally on row 5, column 4 and row 5, column 5.
7. The 30 region at the center needs a total of exactly 30 (3 of its 6 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–5 domino horizontally on row 4, column 2 and row 4, column 3.
8. Row 4, column 1 has only one open neighbor, so its domino must run to row 5, column 1; the 30 region at the center needs a total of exactly 30 (5 of its 6 cells are already filled) — nothing else fits. Place the 3–5 domino vertically on row 4, column 1 and row 5, column 1.
9. Row 5, column 2 has only one open neighbor, so its domino must run to row 5, column 3; the “=” region at the bottom-left needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 0–3 domino horizontally on row 5, column 2 and row 5, column 3.

Start with the 16 region at the center: the only way to make it with these dominoes is 4 + 6 + 6.

• the “<4” region at the top-left: 1, 1
• the “=” region at the top: 4, 4, 4
• the 9 region at the top-right: 3, 6
• the “=” region at the left: 2, 2, 2
• the 16 region at the center: 4, 6, 6
• the “=” region at the bottom-left: 0, 0, 0, 0
• the single-cell “>4” region at the bottom: 5

1. The 16 region at the center needs a total of exactly 16 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 6–6 domino horizontally on row 4, column 3 and row 4, column 4.
2. The “=” region at the bottom-left 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 0–5 domino vertically on row 6, column 2 and row 7, column 2.
3. The “=” region at the bottom-left needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 0–4 domino vertically on row 4, column 2 and row 5, column 2.
4. The “=” region at the bottom-left needs every pip to match (2 of its 4 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–2 domino vertically on row 4, column 1 and row 5, column 1.
5. The “=” region at the left needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 1–2 domino vertically on row 2, column 1 and row 3, column 1.
6. The “=” region at the top needs every pip to match — nothing else fits. Place the 2–4 domino vertically on row 2, column 2 and row 3, column 2.
7. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the “<4” region at the top-left must total less than 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–4 domino horizontally on row 1, column 1 and row 1, column 2.
8. The 9 region at the top-right needs a total of exactly 9 — nothing else fits. Place the 4–6 domino horizontally on row 1, column 3 and row 1, column 4.
9. Row 2, column 4 has only one open neighbor, so its domino must run to row 2, column 3; the 9 region at the top-right needs a total of exactly 9 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–3 domino horizontally on row 2, column 3 and row 2, column 4.
10. Row 6, column 1 has only one open neighbor, so its domino must run to row 7, column 1; the “=” region at the bottom-left needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 0–1 domino vertically on row 6, column 1 and row 7, column 1.