Pips answers for Monday, June 1, 2026

Start with the 3 region at the top: the only way to make it with these dominoes is 1 + 1 + 1.

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

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

Start with the “≠” region at the top-right: the only way to make it with these dominoes is 0 + 1 + 3 + 4 + 5 + 6.

• the “≠” region at the top-right: 0, 1, 3, 4, 5, 6
• the 2 region at the left: 1, 1
• the 10 region at the bottom-left: 5, 5
• the 12 region at the bottom-right: 6, 6
• the 2 region at the bottom: 1, 1

1. The 2 region at the bottom needs a total of exactly 2 — 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 horizontally on row 5, column 3 and row 5, column 4.
2. The “≠” region at the top-right needs every pip to be different — nothing else fits. Place the 6–6 domino vertically on row 3, column 4 and row 4, column 4.
3. The 10 region at the bottom-left needs a total of exactly 10 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–5 domino horizontally on row 4, column 1 and row 4, column 2.
4. The 2 region at the left needs a total of exactly 2 — nothing else fits. Place the 1–1 domino vertically on row 2, column 1 and row 3, column 1.
5. Row 5, column 2 has only one open neighbor, so its domino must run to row 5, column 1; the 2 region at the bottom needs a total of exactly 2 (1 of its 2 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 0–1 domino horizontally on row 5, column 1 and row 5, column 2.
6. The “≠” region at the top-right needs every pip to be different (1 of its 6 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 3–3 domino horizontally on row 1, column 1 and row 1, column 2.
7. The “≠” region at the top-right needs every pip to be different (2 of its 6 cells are already filled) — 8 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–5 domino horizontally on row 1, column 3 and row 1, column 4.
8. Row 2, column 3 has only one open neighbor, so its domino must run to row 2, column 4; the “≠” region at the top-right needs every pip to be different (4 of its 6 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 0–4 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 0 region at the bottom: the only way to make it with these dominoes is 0 + 0 + 0.

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

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