Pips answers for Wednesday, November 12, 2025

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

• the single-cell 3 region at the top-left: 3
• the 30 region at the center: 5, 5, 5, 5, 5, 5
• the single-cell 0 region at the top-right: 0
• the single-cell “<2” region at the bottom-left: 1

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

Start with the 4 region at the right: only 0 + 0 + 4, or 0 + 2 + 2 can make it with these dominoes.

• the 6 region at the top-left: 3, 3
• the “=” region at the top: 6, 6
• the 8 region at the top-right: 4, 4
• the 4 region at the right: 0, 2, 2
• the “=” region at the bottom: 5, 5, 5

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

Start with the “=” region at the top-left: all 4 cells must show the same number, and only 1 or 2 or 3 appear on enough dominoes to fill it.

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

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