Pips answers for Tuesday, March 24, 2026

Start with the 6 region at the top: only 1 + 5, or 2 + 4, or 3 + 3 can make it with these dominoes.

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

1. The 6 region at the bottom-left 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 4–5 domino vertically on row 2, column 1 and row 3, column 1.
2. The 6 region at the bottom-left needs a total of exactly 6 (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 1–2 domino horizontally on row 3, column 2 and row 3, column 3.
3. The single-cell 0 region at the right needs a total of exactly 0 — nothing else fits. Place the 0–3 domino vertically on row 2, column 4 and row 3, column 4.
4. The 6 region at the center needs a total of exactly 6 (1 of its 3 cells is already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–2 domino horizontally on row 2, column 2 and row 2, column 3.
5. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the 6 region at the top needs a total of exactly 6 — nothing else fits. Place the 3–3 domino horizontally on row 1, column 2 and row 1, column 3.

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

• the single-cell “<5” region at the top-left: 0
• the “=” region at the top: 5, 5
• the 7 region at the left: 3, 4
• the 10 region at the right: 4, 6
• the 7 region at the bottom-left: 3, 4
• the “>10” region at the bottom-right: 5, 6
• the “=” region at the bottom: 1, 1

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

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

• the 9 region at the top-left: 4, 5
• the 5 region at the top: 2, 3
• the 3 region at the right: 0, 1, 1, 1
• the 11 region at the left: 5, 6
• the “=” region at the center: 4, 4
• the “=” region at the bottom-left: 0, 0, 0, 0, 0
• the single-cell 4 region at the left: 4
• the “=” region at the bottom: 2, 2, 2, 2
• the “=” region at the bottom-left: 3, 3
• the 7 region at the bottom-right: 3, 4

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