Pips answers for Thursday, November 20, 2025

Start with the “=” region at the center: all 3 cells must show the same number, and only 0 or 5 appear on enough dominoes to fill it.

• the single-cell 5 region at the top-left: 5
• the “=” region at the center: 0, 0, 0
• the “<8” region at the top: 2, 5
• the 3 region at the bottom-right: 1, 2

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

Start with the “≠” region at the center: the only way to make it with these dominoes is 0 + 2 + 3 + 4 + 5 + 6.

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

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

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

Start with the 3 region at the bottom-left: the only way to make it with these dominoes is 0 + 1 + 1 + 1.

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

1. The 18 region at the top-left needs a total of exactly 18 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino vertically on row 2, column 1 and row 3, column 1.
2. The 18 region at the top-left needs a total of exactly 18 (1 of its 3 cells is already filled) — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–6 domino horizontally on row 1, column 2 and row 1, column 3.
3. The single-cell 3 region at the bottom-right needs a total of exactly 3 — 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 4, column 6 and row 5, column 6.
4. The “=” region at the right needs every pip to match (1 of its 4 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 2, column 7 and row 3, column 7.
5. The “=” region at the top-right needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 1, column 6 and row 1, column 7.
6. The “=” region at the top-right needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 5–6 domino horizontally on row 2, column 5 and row 2, column 6.
7. Row 3, column 5 has only one open neighbor, so its domino must run to row 3, column 6; the “=” region at the center needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 2–6 domino horizontally on row 3, column 5 and row 3, column 6.
8. The 18 region at the top-left needs a total of exactly 18 (2 of its 3 cells are already filled) — nothing else fits. Place the 4–6 domino horizontally on row 2, column 2 and row 2, column 3.
9. Row 4, column 7 has only one open neighbor, so its domino must run to row 5, column 7; the “=” region at the right needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 1–2 domino vertically on row 4, column 7 and row 5, column 7.
10. The 3 region at the bottom-left needs a total of exactly 3 (1 of its 4 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 1–1 domino vertically on row 3, column 2 and row 4, column 2.
11. Row 3, column 3 has only one open neighbor, so its domino must run to row 4, column 3; 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 vertically on row 3, column 3 and row 4, column 3.
12. Row 5, column 2 has only one open neighbor, so its domino must run to row 5, column 3; the 3 region at the bottom-left needs a total of exactly 3 (3 of its 4 cells are already filled) — nothing else fits. Place the 0–4 domino horizontally on row 5, column 2 and row 5, column 3.

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