Pips answers for Thursday, November 13, 2025

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

• the single-cell 3 region at the top-left: 3
• the 11 region at the top-right: 5, 6
• the 15 region at the bottom-right: 5, 5, 5

1. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the single-cell 3 region at the top-left needs a total of exactly 3 — nothing else fits. Place the 0–3 domino horizontally on row 1, column 1 and row 1, column 2.
2. The 15 region at the bottom-right 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 vertically on row 3, column 3 and row 4, column 3.
3. The 11 region at the top-right needs a total of exactly 11 — nothing else fits. Place the 5–6 domino vertically on row 2, column 4 and row 3, column 4.
4. Row 1, column 4 has only one open neighbor, so its domino must run to row 1, column 3; the 11 region at the top-right needs a total of exactly 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–5 domino horizontally on row 1, column 3 and row 1, column 4.

Start with the “=” region at the bottom-right: all 4 cells must show the same number, and only 4s appear on enough dominoes to fill it.

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

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

Start with the 3 region at the top: only 0 + 3, or 1 + 2 can make it with these dominoes.

• the 3 region at the top: 1, 2
• the 10 region at the top-right: 4, 6
• the “=” region at the top-left: 4, 4
• the “<2” region at the left: 0, 0
• the “>14” region at the center: 6, 6, 6
• the 6 region at the right: 3, 3
• the single-cell “<3” region at the center: 2
• the “≠” region at the bottom: 1, 2, 3, 5
• the single-cell “>2” region at the bottom-left: 5
• the single-cell “<2” region at the bottom-right: 1

1. The “≠” region at the bottom needs every pip to be different — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–1 domino horizontally on row 7, column 4 and row 7, column 5.
2. The “<2” region at the left must total less than 2 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–4 domino vertically on row 2, column 1 and row 3, column 1.
3. The “=” region at the top-left needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 2–4 domino horizontally on row 2, column 2 and row 2, column 3.
4. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the 3 region at the top needs a total of exactly 3 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–6 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 10 region at the top-right needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino vertically on row 2, column 4 and row 3, column 4.
6. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 3; the “<2” region at the left must total less than 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–6 domino horizontally on row 3, column 2 and row 3, column 3.
7. The single-cell “<3” region at the center must total less than 3 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–2 domino vertically on row 5, column 3 and row 6, column 3.
8. Row 6, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the “≠” region at the bottom needs every pip to be different (2 of its 4 cells are already filled) — 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 5, column 4 and row 6, column 4.
9. Row 4, column 3 has only one open neighbor, so its domino must run to row 4, column 4; the “>14” region at the center must total more than 14 (2 of its 3 cells are already filled) — nothing else fits. Place the 3–6 domino horizontally on row 4, column 3 and row 4, column 4.
10. Row 7, column 2 has only one open neighbor, so its domino must run to row 7, column 3; the single-cell “>2” region at the bottom-left must total more than 2 — nothing else fits. Place the 5–5 domino horizontally on row 7, column 2 and row 7, column 3.

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