Pips answers for Sunday, November 23, 2025

Start with the 5 region at the top-left: only 0 + 5, or 1 + 4 can make it with these dominoes.

• the 5 region at the top-left: 0, 5
• the 6 region at the bottom-left: 1, 5
• the 5 region at the bottom-right: 1, 4
• the single-cell 6 region at the bottom: 6

1. The 6 region at the bottom-left needs a total of exactly 6 — nothing else fits. Place the 1–6 domino horizontally on row 3, column 1 and row 3, column 2.
2. The 5 region at the top-left needs a total of exactly 5 — nothing else fits. Place the 5–5 domino vertically on row 1, column 1 and row 2, column 1.
3. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the 5 region at the top-left needs a total of exactly 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–0 domino horizontally on row 1, column 2 and row 1, column 3.
4. Row 2, column 3 has only one open neighbor, so its domino must run to row 3, column 3; the 5 region at the bottom-right needs a total of exactly 5 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–4 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 “=” region at the left: all 4 cells must show the same number, and only 1 or 2 appear on enough dominoes to fill it.

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

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

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

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

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