Pips answers for Monday, August 18, 2025

Start with the “=” region at the top: all 4 cells must show the same number, and only 2s appear on enough dominoes to fill it.

• the “=” region at the top: 2, 2, 2, 2
• the single-cell 5 region at the top-right: 5
• the 12 region at the bottom: 6, 6

1. The “=” region at the top needs every pip to match — nothing else fits. Place the 2–5 domino horizontally on row 1, column 3 and row 1, column 4.
2. The “=” region at the top needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 2–2 domino horizontally on row 2, column 2 and row 2, column 3.
3. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 1; the “=” region at the top needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 2–3 domino horizontally on row 1, column 1 and row 1, column 2.
4. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 3; the 12 region at the bottom needs a total of exactly 12 — nothing else fits. Place the 6–6 domino horizontally on row 3, column 2 and row 3, column 3.

Start with the 18 region at the bottom: the only way to make it with these dominoes is 6 + 6 + 6.

• the “<13” region at the left: 3, 4, 4
• the 18 region at the bottom: 6, 6, 6
• the single-cell “<3” region at the right: 1
• the single-cell 2 region at the bottom-left: 2
• the single-cell “>4” region at the bottom-right: 5

1. The 18 region at the bottom needs a total of exactly 18 — nothing else fits. Place the 2–6 domino horizontally on row 3, column 2 and row 3, column 3.
2. The 18 region at the bottom needs a total of exactly 18 (1 of its 3 cells is already filled) — nothing else fits. Place the 1–6 domino horizontally on row 2, column 4 and row 2, column 5.
3. Row 3, column 4 has only one open neighbor, so its domino must run to row 3, column 5; the 18 region at the bottom needs a total of exactly 18 (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.
4. The “<13” region at the left must total less than 13 — 6 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–6 domino horizontally on row 1, column 1 and row 1, column 2.
5. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the “<13” region at the left must total less than 13 (1 of its 3 cells is already filled) — nothing else fits. Place the 4–4 domino horizontally on row 2, column 1 and row 2, column 2.

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

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

• the “<3” region at the top-left: 1, 1
• the “=” region at the top-right: 3, 3
• the “=” region at the left: 6, 6
• the single-cell 1 region at the left: 1
• the “=” region at the bottom: 2, 2
• the 18 region at the right: 6, 6, 6
• the 10 region at the bottom-right: 5, 5
• the 0 region at the bottom-left: 0, 0, 0, 0
• the single-cell 4 region at the bottom-right: 4

1. The 18 region at the right needs a total of exactly 18 — nothing else fits. Place the 6–6 domino vertically on row 2, column 7 and row 3, column 7.
2. The 10 region at the bottom-right needs a total of exactly 10 — nothing else fits. Place the 4–5 domino horizontally on row 3, column 8 and row 3, column 9.
3. The “=” region at the top-right needs every pip to match — nothing else fits. Place the 3–5 domino horizontally on row 2, column 8 and row 2, column 9.
4. Row 1, column 8 has only one open neighbor, so its domino must run to row 1, column 7; the “=” region at the top-right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 1–3 domino horizontally on row 1, column 7 and row 1, column 8.
5. The 0 region at the bottom-left 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 0–0 domino horizontally on row 3, column 2 and row 3, column 3.
6. Row 3, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 0 region at the bottom-left needs a total of exactly 0 (2 of its 4 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–6 domino vertically on row 2, column 1 and row 3, column 1.
7. Row 3, column 4 has only one open neighbor, so its domino must run to row 3, column 5; the 0 region at the bottom-left needs a total of exactly 0 (3 of its 4 cells are already filled) — nothing else fits. Place the 0–2 domino horizontally on row 3, column 4 and row 3, column 5.
8. Row 2, column 5 has only one open neighbor, so its domino must run to row 2, column 6; 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 2, column 5 and row 2, column 6.
9. The “<3” region at the top-left must total less than 3 — 2 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.
10. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, 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 1–6 domino horizontally on row 2, column 2 and row 2, column 3.