Pips answers for Tuesday, November 25, 2025

Start with the 16 region at the right: the only way to make it with these dominoes is 5 + 5 + 6.

• the 7 region at the left: 2, 5
• the 16 region at the right: 5, 5, 6
• the single-cell 1 region at the bottom-left: 1
• the 7 region at the bottom: 3, 4

1. The 7 region at the left needs a total of exactly 7 — nothing else fits. Place the 1–2 domino vertically on row 3, column 1 and row 4, column 1.
2. The 16 region at the right needs a total of exactly 16 — nothing else fits. Place the 5–5 domino vertically on row 3, column 4 and row 4, column 4.
3. Row 2, column 1 has only one open neighbor, so its domino must run to row 1, column 1; the 7 region at the left needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–5 domino vertically on row 1, column 1 and row 2, column 1.
4. Row 2, column 4 has only one open neighbor, so its domino must run to row 1, column 4; the 16 region at the right needs a total of exactly 16 (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.
5. Row 4, column 2 has only one open neighbor, so its domino must run to row 4, column 3; the 7 region at the bottom needs a total of exactly 7 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–4 domino horizontally on row 4, column 2 and row 4, column 3.

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

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

• the “=” region at the left: 2, 2, 2, 2
• the “=” region at the top: 1, 1
• the 7 region at the right: 1, 3, 3
• the “=” region at the bottom: 5, 5, 5

1. The 7 region at the right needs a total of exactly 7 — 13 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 1, column 4 and row 1, column 5.
2. The “=” region at the left 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 1–2 domino horizontally on row 1, column 2 and row 1, column 3.
3. 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 2–2 domino horizontally on row 2, column 1 and row 2, 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 left needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 2–5 domino vertically on row 2, column 3 and row 3, column 3.
5. Row 3, column 4 has only one open neighbor, so its domino must run to row 3, column 5; the “=” region at the bottom needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 3–5 domino horizontally on row 3, column 4 and row 3, column 5.
6. Row 2, column 5 has only one open neighbor, so its domino must run to row 2, column 6; the 7 region at the right needs a total of exactly 7 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–4 domino horizontally on row 2, column 5 and row 2, column 6.
7. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 1; the “=” region at the bottom needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 5–6 domino horizontally on row 3, column 1 and row 3, column 2.

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

• the single-cell “>2” region at the top-left: 4
• the 10 region at the top-left: 4, 6
• the “=” region at the center: 1, 1, 1
• the 17 region at the right: 3, 4, 5, 5
• the 0 region at the top-right: 0, 0
• the 10 region at the left: 5, 5
• the 5 region at the right: 2, 3
• the 0 region at the bottom-left: 0, 0, 0
• the 12 region at the bottom-right: 6, 6

1. The 0 region at the bottom-left needs a total of exactly 0 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–0 domino horizontally on row 4, column 3 and row 4, column 4.
2. The “=” region at the center 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 1–1 domino vertically on row 1, column 5 and row 2, column 5.
3. Row 3, column 5 has only one open neighbor, so its domino must run to row 4, column 5; the “=” region at the center needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 1–2 domino vertically on row 3, column 5 and row 4, column 5.
4. Row 3, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the 0 region at the bottom-left needs a total of exactly 0 (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 0–6 domino vertically on row 2, column 3 and row 3, column 3.
5. The 0 region at the top-right needs a total of exactly 0 — 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 1, column 7 and row 1, column 8.
6. Row 1, column 9 has only one open neighbor, so its domino must run to row 2, column 9; the 0 region at the top-right needs a total of exactly 0 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–3 domino vertically on row 1, column 9 and row 2, column 9.
7. Row 3, column 9 has only one open neighbor, so its domino must run to row 4, column 9; the 5 region at the right needs a total of exactly 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–6 domino vertically on row 3, column 9 and row 4, column 9.
8. The 17 region at the right needs a total of exactly 17 (1 of its 4 cells is already filled) — nothing else fits. Place the 3–6 domino horizontally on row 4, column 7 and row 4, column 8.
9. Row 2, column 7 has only one open neighbor, so its domino must run to row 3, column 7; the 17 region at the right needs a total of exactly 17 (2 of its 4 cells are already filled) — nothing else fits. Place the 5–5 domino vertically on row 2, column 7 and row 3, column 7.
10. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the 10 region at the top-left needs a total of exactly 10 (1 of its 2 cells is 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 1, column 3 and row 1, column 4.
11. The 10 region at the left needs a total of exactly 10 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–5 domino vertically on row 1, column 1 and row 2, column 1.
12. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; the 10 region at the left needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–5 domino vertically on row 3, column 1 and row 4, column 1.