Pips answers for Monday, July 6, 2026

Start with the “<3” region at the bottom-left: the only way to make it with these dominoes is 1 + 1.

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

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

Start with the 15 region at the right: only 3 + 6 + 6, or 4 + 5 + 6, or 5 + 5 + 5 can make it with these dominoes.

• the “=” region at the top-left: 4, 4
• the 15 region at the right: 3, 6, 6
• the single-cell 4 region at the left: 4
• the 6 region at the bottom-left: 0, 6
• the single-cell 3 region at the center: 3
• the “=” region at the bottom: 5, 5, 5
• the single-cell 4 region at the right: 4

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

Start with the single-cell 3 region at the top-left: the only way to make it with these dominoes is 3.

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

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