Pips answers for Sunday, May 24, 2026

Start with the 10 region at the left: the only way to make it with these dominoes is 0 + 5 + 5.

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

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

Start with the 15 region at the bottom-left: only 4 + 5 + 6, or 5 + 5 + 5 can make it with these dominoes.

• the “=” region at the top-right: 2, 2
• the single-cell “>4” region at the right: 5
• the single-cell “<2” region at the left: 1
• the “=” region at the left: 4, 4
• the 13 region at the right: 1, 6, 6
• the 15 region at the bottom-left: 5, 5, 5

1. The single-cell “<2” region at the left must total less than 2 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–4 domino horizontally on row 3, column 1 and row 3, column 2.
2. The “=” region at the left needs every pip to match (1 of its 2 cells is already filled) — 3 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 3 and row 4, column 3.
3. Row 4, column 1 has only one open neighbor, so its domino must run to row 4, column 2; the 15 region at the bottom-left needs a total of exactly 15 (1 of its 3 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 4, column 1 and row 4, column 2.
4. The 13 region at the right needs a total of exactly 13 — 5 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 3, column 5 and row 3, column 6.
5. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 13 region at the right needs a total of exactly 13 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–1 domino vertically on row 3, column 4 and row 4, column 4.
6. The “=” region at the top-right needs every pip to match — nothing else fits. Place the 2–5 domino horizontally on row 2, column 5 and row 2, column 6.
7. Row 1, column 6 has only one open neighbor, so its domino must run to row 1, 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 2–4 domino horizontally on row 1, column 5 and row 1, column 6.

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

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

• the single-cell 1 region at the top-left: 1
• the single-cell 1 region at the top-left: 1
• the 11 region at the top: 2, 4, 5
• the single-cell 1 region at the top-left: 1
• the 11 region at the top-left: 5, 6
• the “>11” region at the right: 6, 6
• the single-cell “>1” region at the right: 2
• the single-cell 1 region at the left: 1
• the 11 region at the bottom-right: 0, 2, 4, 5
• the “=” region at the bottom-left: 3, 3, 3, 3
• the single-cell 1 region at the bottom: 1
• the “=” region at the bottom-left: 4, 4
• the 11 region at the bottom: 5, 6
• the single-cell 1 region at the bottom-left: 1
• the single-cell 1 region at the bottom-right: 1
• the single-cell 1 region at the bottom-right: 1
• the single-cell “>1” region at the bottom-right: 2

1. The single-cell 1 region at the top-left needs a total of exactly 1 — 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 horizontally on row 1, column 2 and row 1, column 3.
2. The 11 region at the top-left needs a total of exactly 11 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino vertically on row 2, column 2 and row 3, column 2.
3. The 11 region at the bottom-right needs a total of exactly 11 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–6 domino vertically on row 4, column 6 and row 5, column 6.
4. The 11 region at the bottom needs a total of exactly 11 — nothing else fits. Place the 1–5 domino horizontally on row 8, column 3 and row 8, column 4.
5. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the 11 region at the top-left needs a total of exactly 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 3, column 3 and row 3, column 4.
6. The “=” region at the bottom-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 3–3 domino vertically on row 6, column 2 and row 7, column 2.
7. Row 5, column 1 has only one open neighbor, so its domino must run to row 6, column 1; the single-cell 1 region at the left needs a total of exactly 1 — nothing else fits. Place the 1–3 domino vertically on row 5, column 1 and row 6, column 1.
8. Row 7, column 1 has only one open neighbor, so its domino must run to row 8, column 1; the “=” region at the bottom-left needs every pip to match — nothing else fits. Place the 4–4 domino vertically on row 7, column 1 and row 8, column 1.
9. Row 7, column 3 has only one open neighbor, so its domino must run to row 7, column 4; the “=” region at the bottom-left needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 3–6 domino horizontally on row 7, column 3 and row 7, column 4.
10. Row 4, column 7 has only one open neighbor, so its domino must run to row 4, column 8; the “>11” region at the right must total more than 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–6 domino horizontally on row 4, column 7 and row 4, column 8.
11. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the 11 region at the top needs a total of exactly 11 (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 vertically on row 1, column 4 and row 2, column 4.
12. The single-cell 1 region at the bottom-right needs a total of exactly 1 — 2 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 8, column 7 and row 8, column 8.
13. The 11 region at the bottom-right needs a total of exactly 11 (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 1–4 domino vertically on row 7, column 6 and row 8, column 6.
14. Row 6, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the single-cell 1 region at the bottom needs a total of exactly 1 — nothing else fits. Place the 0–1 domino vertically on row 5, column 4 and row 6, column 4.
15. Row 6, column 6 has only one open neighbor, so its domino must run to row 6, column 7; the 11 region at the bottom-right needs a total of exactly 11 (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 horizontally on row 6, column 6 and row 6, column 7.

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