Pips answers for Friday, November 21, 2025

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

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

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

Start with the 12 region at the top: the only way to make it with these dominoes is 6 + 6.

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

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

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 13 region at the top-right: 1, 6, 6
• the 13 region at the top-left: 1, 6, 6
• the “=” region at the right: 3, 3, 3
• the 13 region at the bottom-left: 1, 2, 4, 6
• the 13 region at the bottom: 3, 4, 6
• the single-cell “>4” region at the right: 5
• the “=” region at the bottom-right: 4, 4, 4

1. Row 5, column 5 has only one open neighbor, so its domino must run to row 5, column 6; the “=” region at the bottom-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 4–4 domino horizontally on row 5, column 5 and row 5, column 6.
2. The 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 0–1 domino vertically on row 1, column 2 and row 2, column 2.
3. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the 0 region at the top-left needs a total of exactly 0 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–6 domino vertically on row 1, column 3 and row 2, column 3.
4. 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 3–4 domino vertically on row 4, column 7 and row 5, column 7.
5. The “=” region at the right needs every pip to match (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 3–5 domino horizontally on row 3, column 6 and row 3, column 7.
6. The 13 region at the top-right needs a total of exactly 13 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–3 domino vertically on row 1, column 7 and row 2, column 7.
7. Row 1, column 5 has only one open neighbor, so its domino must run to row 1, column 6; the 13 region at the top-right needs a total of exactly 13 (1 of its 3 cells is already filled) — nothing else fits. Place the 6–6 domino horizontally on row 1, column 5 and row 1, column 6.
8. The 13 region at the bottom-left needs a total of exactly 13 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–6 domino vertically on row 2, column 1 and row 3, column 1.
9. The 13 region at the bottom needs a total of exactly 13 — 10 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 2 and row 3, column 3.
10. The 13 region at the bottom-left needs a total of exactly 13 (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 3–6 domino horizontally on row 4, column 2 and row 4, column 3.
11. Row 5, column 2 has only one open neighbor, so its domino must run to row 5, column 3; the 13 region at the bottom-left needs a total of exactly 13 (3 of its 4 cells are already filled) — nothing else fits. Place the 1–4 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.