Pips answers for Wednesday, February 25, 2026

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: 6
• the single-cell 2 region at the top-left: 2
• the 30 region at the center: 5, 5, 5, 5, 5, 5
• the single-cell “<2” region at the bottom: 0

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

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

• the single-cell “>0” region at the top-left: 2
• the 15 region at the top: 4, 5, 6
• the 15 region at the center: 5, 5, 5
• the “=” region at the bottom: 3, 3, 3, 3, 3
• the single-cell “<6” region at the bottom-left: 1

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 3–3 domino horizontally on row 5, column 3 and row 5, column 4.
2. Row 5, column 2 has only one open neighbor, so its domino must run to row 5, column 1; the “=” region at the bottom 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 horizontally on row 5, column 1 and row 5, column 2.
3. Row 5, column 5 has only one open neighbor, so its domino must run to row 5, column 6; the “=” region at the bottom needs every pip to match (3 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 3–6 domino horizontally on row 5, column 5 and row 5, column 6.
4. The “=” region at the bottom needs every pip to match (4 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 3–5 domino vertically on row 3, column 3 and row 4, column 3.
5. The 15 region at the center needs a total of exactly 15 (1 of its 3 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 2, column 3 and row 2, column 4.
6. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 15 region at the center needs a total of exactly 15 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–5 domino vertically on row 3, column 4 and row 4, column 4.
7. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the single-cell “>0” region at the top-left must total more than 0 — nothing else fits. Place the 2–4 domino horizontally on row 1, column 2 and row 1, column 3.
8. Row 1, column 4 has only one open neighbor, so its domino must run to row 1, column 5; the 15 region at the top needs a total of exactly 15 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–6 domino horizontally on row 1, column 4 and row 1, column 5.

Start with the 2 region at the top-right: only 0 + 2, or 1 + 1 can make it with these dominoes.

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

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