Pips answers for Tuesday, September 16, 2025

Start with the “=” region at the top-right: all 4 cells must show the same number, and only 1s appear on enough dominoes to fill it.

• the “=” region at the top-right: 1, 1, 1, 1
• the 11 region at the bottom-right: 5, 6
• the 4 region at the bottom: 2, 2

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

Start with the 4 region at the top-right: the only way to make it with these dominoes is 0 + 2 + 2.

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

1. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the 4 region at the top-right needs a total of exactly 4 — nothing else fits. Place the 2–2 domino vertically on row 1, column 3 and row 2, column 3.
2. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the “=” region at the left needs every pip to match — nothing else fits. Place the 6–6 domino horizontally on row 2, column 1 and row 2, column 2.
3. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the 4 region at the top-right needs a total of exactly 4 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–5 domino vertically on row 2, column 4 and row 3, column 4.
4. The “=” region at the left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 3–6 domino vertically on row 3, column 2 and row 4, column 2.
5. Row 3, column 3 has only one open neighbor, so its domino must run to row 4, column 3; the “=” region at the right needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 3–5 domino vertically on row 3, column 3 and row 4, column 3.
6. Row 4, column 1 has only one open neighbor, so its domino must run to row 5, 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 3–4 domino vertically on row 4, column 1 and row 5, column 1.
7. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the “=” region at the right needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 2–5 domino vertically on row 4, column 4 and row 5, column 4.

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

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

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

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