Pips answers for Friday, September 5, 2025

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

• the single-cell 3 region at the top: 3
• the “=” region at the center: 3, 3, 3
• the 7 region at the bottom-left: 2, 5
• the single-cell 3 region at the bottom: 3

1. The “=” region at the center needs every pip to match — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–3 domino vertically on row 1, column 2 and row 2, column 2.
2. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the “=” region at the center needs every pip to match (1 of its 3 cells is already filled) — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–5 domino vertically on row 2, column 1 and row 3, column 1.
3. Row 4, column 1 has only one open neighbor, so its domino must run to row 4, column 2; the 7 region at the bottom-left needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–3 domino horizontally on row 4, column 1 and row 4, column 2.
4. Row 2, column 3 has only one open neighbor, so its domino must run to row 3, column 3; the “=” region at the center needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 1–3 domino vertically on row 2, column 3 and row 3, column 3.

This board has more than one valid layout; the steps follow the official answer.

Start with the “<4” region at the top-left: the only way to make it with these dominoes is 1 + 2.

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

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

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

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

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