Pips answers for Thursday, September 18, 2025

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

• the 6 region at the bottom-left: 0, 0, 6
• the “=” region at the top-right: 3, 3
• the 0 region at the bottom: 0, 0

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

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

Start with the 20 region at the center: the only way to make it with these dominoes is 4 + 5 + 5 + 6.

• the 4 region at the top-left: 0, 4
• the “≠” region at the left: 0, 3
• the 20 region at the center: 4, 5, 5, 6
• the “=” region at the right: 2, 2, 2
• the single-cell “<3” region at the bottom: 0

1. The “=” region at the right needs every pip to match — 6 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 4, column 3 and row 4, column 4.
2. 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 2–4 domino horizontally on row 2, column 3 and row 2, column 4.
3. The 20 region at the center needs a total of exactly 20 (1 of its 4 cells is already filled) — nothing else fits. Place the 2–6 domino horizontally on row 3, column 3 and row 3, column 4.
4. The 4 region at the top-left needs a total of exactly 4 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–5 domino vertically on row 1, column 2 and row 2, column 2.
5. The “≠” region at the left needs every pip to be different — nothing else fits. Place the 0–5 domino horizontally on row 3, column 1 and row 3, column 2.
6. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 4 region at the top-left needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–3 domino vertically on row 1, column 1 and row 2, column 1.

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

Start with the “=” region at the left: all 5 cells must show the same number, and only 2s appear on enough dominoes to fill it.

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

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