Pips answers for Friday, April 24, 2026

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

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

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

Start with the 7 region at the right: only 1 + 6, or 2 + 5 can make it with these dominoes.

• the 7 region at the top-left: 1, 3, 3
• the 7 region at the left: 1, 1, 5
• the single-cell “<2” region at the top-right: 0
• the single-cell “<5” region at the left: 3
• the 7 region at the right: 1, 6
• the 7 region at the bottom-right: 2, 5
• the single-cell “<3” region at the bottom-right: 2

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

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

Start with the “<2” region at the center: only 0 + 0 + 0, or 0 + 0 + 1 can make it with these dominoes.

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

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