Pips answers for Friday, September 25, 2026

Start with the “=” region at the right: all 3 cells must show the same number, and only 4 or 6 appear on enough dominoes to fill it.

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

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

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

Start with the 7 region at the top: only 1 + 6, or 3 + 4 can make it with these dominoes.

• the single-cell 2 region at the top-left: 2
• the 7 region at the top: 1, 6
• the 7 region at the top-right: 1, 6
• the 7 region at the bottom-left: 0, 1, 2, 4
• the 7 region at the right: 0, 2, 2, 3
• the single-cell 3 region at the center: 3

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

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

Start with the 18 region at the top-left: the only way to make it with these dominoes is 6 + 6 + 6.

• the single-cell 0 region at the top-left: 0
• the 14 region at the top: 2, 2, 5, 5
• the 18 region at the top-left: 6, 6, 6
• the 14 region at the top-right: 4, 4, 6
• the single-cell 1 region at the top-right: 1
• the 12 region at the right: 3, 3, 6
• the single-cell “<3” region at the left: 2
• the single-cell 0 region at the center: 0
• the single-cell 4 region at the right: 4
• the 9 region at the bottom-left: 4, 5
• the 6 region at the bottom: 3, 3
• the 0 region at the bottom-right: 0, 0
• the single-cell “>0” region at the bottom-right: 1
• the 0 region at the bottom-left: 0, 0
• the 10 region at the bottom: 5, 5
• the single-cell 0 region at the bottom-right: 0
• the “=” region at the bottom-right: 2, 2

1. The 14 region at the top-right needs a total of exactly 14 — nothing else fits. Place the 1–6 domino horizontally on row 2, column 7 and row 2, column 8.
2. The 18 region at the top-left needs a total of exactly 18 — nothing else fits. Place the 6–6 domino vertically on row 3, column 3 and row 4, column 3.
3. The 0 region at the bottom-right needs a total of exactly 0 — nothing else fits. Place the 0–0 domino vertically on row 9, column 6 and row 10, column 6.
4. The 14 region at the top needs a total of exactly 14 — 2 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 3 and row 2, column 3.
5. The 0 region at the bottom-left needs a total of exactly 0 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–5 domino horizontally on row 10, column 2 and row 10, column 3.
6. The 9 region at the bottom-left needs a total of exactly 9 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–4 domino vertically on row 9, column 1 and row 10, column 1.
7. Row 7, column 1 has only one open neighbor, so its domino must run to row 8, column 1; the single-cell “<3” region at the left must total less than 3 — nothing else fits. Place the 2–5 domino vertically on row 7, column 1 and row 8, column 1.
8. The 12 region at the right 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 vertically on row 6, column 6 and row 7, column 6.
9. Row 6, column 7 has only one open neighbor, so its domino must run to row 6, column 8; the 12 region at the right needs a total of exactly 12 (1 of its 3 cells is already filled) — nothing else fits. Place the 3–3 domino horizontally on row 6, column 7 and row 6, column 8.
10. The 6 region at the bottom needs a total of exactly 6 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–5 domino vertically on row 9, column 4 and row 10, column 4.
11. Row 3, column 7 has only one open neighbor, so its domino must run to row 4, column 7; the 14 region at the top-right needs a total of exactly 14 (1 of its 3 cells is already filled) — nothing else fits. Place the 4–4 domino vertically on row 3, column 7 and row 4, column 7.
12. Row 1, column 4 has only one open neighbor, so its domino must run to row 1, column 5; the 14 region at the top needs a total of exactly 14 (1 of its 4 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 1, column 4 and row 1, column 5.
13. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the single-cell 0 region at the top-left needs a total of exactly 0 — nothing else fits. Place the 0–2 domino horizontally on row 1, column 1 and row 1, column 2.
14. Row 7, column 4 has only one open neighbor, so its domino must run to row 8, column 4; the single-cell 0 region at the center needs a total of exactly 0 — nothing else fits. Place the 0–3 domino vertically on row 7, column 4 and row 8, column 4.
15. Row 8, column 6 has only one open neighbor, so its domino must run to row 8, column 7; the 0 region at the bottom-right needs a total of exactly 0 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–1 domino horizontally on row 8, column 6 and row 8, column 7.
16. Row 10, column 7 has only one open neighbor, so its domino must run to row 10, column 8; the “=” region at the bottom-right needs every pip to match — nothing else fits. Place the 2–2 domino horizontally on row 10, column 7 and row 10, column 8.