Pips answers for Wednesday, August 26, 2026

Start with the 28 region at the center: the only way to make it with these dominoes is 5 + 5 + 6 + 6 + 6.

• the 5 region at the top-right: 2, 3
• the 28 region at the center: 5, 5, 6, 6, 6
• the single-cell “<1” region at the left: 0
• the single-cell “<2” region at the bottom: 1

1. The 28 region at the center needs a total of exactly 28 — nothing else fits. Place the 6–6 domino horizontally on row 2, column 2 and row 2, column 3.
2. Row 3, column 1 has only one open neighbor, so its domino must run to row 3, column 2; the single-cell “<1” region at the left must total less than 1 — nothing else fits. Place the 0–6 domino horizontally on row 3, column 1 and row 3, column 2.
3. The 28 region at the center needs a total of exactly 28 (3 of its 5 cells are already filled) — nothing else fits. Place the 1–5 domino vertically on row 3, column 3 and row 4, column 3.
4. The 5 region at the top-right needs a total of exactly 5 — nothing else fits. Place the 2–5 domino vertically on row 2, column 4 and row 3, column 4.
5. Row 1, column 4 has only one open neighbor, so its domino must run to row 1, column 5; the 5 region at the top-right needs a total of exactly 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–4 domino horizontally on row 1, column 4 and row 1, column 5.

Start with the 11 region at the center: the only way to make it with these dominoes is 5 + 6.

• the single-cell “>4” region at the top-left: 5
• the “=” region at the top: 3, 3
• the “=” region at the left: 6, 6, 6
• the “=” region at the center: 1, 1
• the 11 region at the center: 5, 6
• the “=” region at the bottom-right: 5, 5
• the single-cell 4 region at the right: 4
• the single-cell 4 region at the bottom-left: 4
• the single-cell 2 region at the bottom: 2

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

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

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

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

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