Pips answers for Tuesday, October 7, 2025

Start with the 15 region at the bottom: the only way to make it with these dominoes is 5 + 5 + 5.

• the 9 region at the top-left: 3, 3, 3
• the 16 region at the right: 4, 4, 4, 4
• the 15 region at the bottom: 5, 5, 5

1. Row 5, column 1 has only one open neighbor, so its domino must run to row 5, column 2; the 15 region at the bottom needs a total of exactly 15 — 3 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 5, column 1 and row 5, column 2.
2. The 16 region at the right needs a total of exactly 16 — nothing else fits. Place the 4–5 domino vertically on row 4, column 3 and row 5, column 3.
3. The 16 region at the right needs a total of exactly 16 (1 of its 4 cells is already filled) — 4 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 3, column 2 and row 3, column 3.
4. Row 2, column 3 has only one open neighbor, so its domino must run to row 1, column 3; the 16 region at the right needs a total of exactly 16 (3 of its 4 cells are already filled) — nothing else fits. Place the 0–4 domino vertically on row 1, column 3 and row 2, column 3.
5. The 9 region at the top-left needs a total of exactly 9 — 3 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.
6. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the 9 region at the top-left needs a total of exactly 9 (1 of its 3 cells is already filled) — nothing else fits. Place the 3–3 domino vertically on row 2, column 1 and row 3, column 1.

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

• the “<4” region at the top-left: 0, 1
• the “=” region at the top-right: 6, 6, 6
• the “=” region at the left: 2, 2, 2, 2
• the “=” region at the bottom-left: 3, 3, 3
• the 11 region at the bottom-right: 1, 5, 5

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

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

• the 0 region at the top-left: 0, 0
• the 11 region at the left: 3, 4, 4
• the single-cell “<2” region at the top-right: 1
• the “=” region at the right: 6, 6, 6
• the “=” region at the bottom-left: 2, 2
• the 7 region at the bottom: 3, 4
• the “=” region at the bottom: 2, 2
• the 5 region at the bottom-left: 0, 5
• the 11 region at the bottom-right: 0, 5, 6

1. The “=” region at the right needs every pip to match — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 6–6 domino vertically on row 3, column 8 and row 4, column 8.
2. Row 2, column 7 has only one open neighbor, so its domino must run to row 2, column 8; the single-cell “<2” region at the top-right must total less than 2 — nothing else fits. Place the 1–6 domino horizontally on row 2, column 7 and row 2, column 8.
3. The 0 region at the top-left needs a total of exactly 0 — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–0 domino horizontally on row 1, column 1 and row 1, column 2.
4. The 11 region at the left needs a total of exactly 11 — 9 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–4 domino horizontally on row 2, column 1 and row 2, column 2.
5. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; the 11 region at the left needs a total of exactly 11 (2 of its 3 cells are already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–3 domino vertically on row 3, column 1 and row 4, column 1.
6. Row 4, column 2 has only one open neighbor, so its domino must run to row 5, column 2; the “=” region at the bottom-left needs every pip to match (1 of its 2 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 2–5 domino vertically on row 4, column 2 and row 5, column 2.
7. Row 5, column 3 has only one open neighbor, so its domino must run to row 5, column 4; the 5 region at the bottom-left needs a total of exactly 5 (1 of its 2 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–3 domino horizontally on row 5, column 3 and row 5, column 4.
8. Row 4, column 4 has only one open neighbor, so its domino must run to row 4, column 5; the 7 region at the bottom needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–4 domino horizontally on row 4, column 4 and row 4, column 5.
9. Row 4, column 6 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 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–2 domino vertically on row 4, column 6 and row 5, column 6.
10. Row 5, column 7 has only one open neighbor, so its domino must run to row 5, column 8; the 11 region at the bottom-right needs a total of exactly 11 (1 of its 3 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–6 domino horizontally on row 5, column 7 and row 5, column 8.

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