Pips answers for Thursday, August 28, 2025

Start with the 17 region at the bottom-left: only 2 + 5 + 5 + 5, or 4 + 4 + 4 + 5 can make it with these dominoes.

• the “=” region at the top-left: 4, 4, 4
• the 17 region at the bottom-left: 2, 5, 5, 5

1. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the “=” region at the top-left 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 4–4 domino vertically on row 1, column 1 and row 2, column 1.
2. 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 2–4 domino vertically on row 3, column 1 and row 4, column 1.
3. Row 5, column 1 has only one open neighbor, so its domino must run to row 5, column 2; the 17 region at the bottom-left needs a total of exactly 17 (1 of its 4 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 5, column 1 and row 5, column 2.
4. Row 5, column 3 has only one open neighbor, so its domino must run to row 5, column 4; the 17 region at the bottom-left needs a total of exactly 17 (3 of its 4 cells are already filled) — nothing else fits. Place the 1–5 domino horizontally on row 5, column 3 and row 5, column 4.

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

• the 12 region at the top-right: 6, 6
• the “=” region at the right: 1, 1, 1, 1
• the 17 region at the bottom-left: 4, 4, 4, 5

1. The “=” region at the 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 1–1 domino vertically on row 3, column 3 and row 4, column 3.
2. The 12 region at the top-right needs a total of exactly 12 — nothing else fits. Place the 1–6 domino vertically on row 1, column 3 and row 2, column 3.
3. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 1; the 12 region at the top-right needs a total of exactly 12 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino horizontally on row 1, column 1 and row 1, column 2.
4. The 17 region at the bottom-left needs a total of exactly 17 — nothing else fits. Place the 1–4 domino horizontally on row 5, column 2 and row 5, column 3.
5. The 17 region at the bottom-left needs a total of exactly 17 (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 4–5 domino vertically on row 4, column 1 and row 5, column 1.
6. Row 3, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 17 region at the bottom-left needs a total of exactly 17 (3 of its 4 cells are already filled) — nothing else fits. Place the 2–4 domino vertically on row 2, column 1 and row 3, column 1.

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

Start with the 17 region at the left: the only way to make it with these dominoes is 5 + 6 + 6.

• the “=” region at the top-left: 1, 1
• the single-cell “>1” region at the top: 3
• the 7 region at the right: 0, 2, 2, 3
• the “<2” region at the top-right: 0, 1
• the 17 region at the left: 5, 6, 6
• the 9 region at the center: 4, 5
• the single-cell 0 region at the right: 0
• the single-cell 1 region at the bottom-left: 1
• the single-cell 6 region at the bottom-right: 6
• the 0 region at the bottom-right: 0, 0

1. Row 3, column 2 has only one open neighbor, so its domino must run to row 4, column 2; the 17 region at the left needs a total of exactly 17 — nothing else fits. Place the 6–6 domino vertically on row 3, column 2 and row 4, column 2.
2. The 17 region at the left needs a total of exactly 17 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–5 domino vertically on row 4, column 3 and row 5, column 3.
3. The 7 region at the right needs a total of exactly 7 — nothing else fits. Place the 2–6 domino vertically on row 4, column 7 and row 5, column 7.
4. Row 5, column 8 has only one open neighbor, so its domino must run to row 5, column 9; the 0 region at the bottom-right needs a total of exactly 0 — nothing else fits. Place the 0–0 domino horizontally on row 5, column 8 and row 5, column 9.
5. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the “=” region at the top-left needs every pip to match — nothing else fits. Place the 1–1 domino vertically on row 1, column 1 and row 2, column 1.
6. The “<2” region at the top-right must total less than 2 — 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 horizontally on row 1, column 8 and row 1, column 9.
7. The 7 region at the right needs a total of exactly 7 (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 0–3 domino horizontally on row 3, column 7 and row 3, column 8.
8. Row 1, column 7 has only one open neighbor, so its domino must run to row 2, column 7; the 7 region at the right needs a total of exactly 7 (2 of its 4 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 vertically on row 1, column 7 and row 2, column 7.
9. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 9 region at the center needs a total of exactly 9 — 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 3, column 4 and row 4, column 4.
10. Row 1, column 5 has only one open neighbor, so its domino must run to row 2, column 5; the single-cell “>1” region at the top must total more than 1 — 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 1, column 5 and row 2, column 5.

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