Pips answers for Monday, November 24, 2025

Start with the 12 region at the top: the only way to make it with these dominoes is 6 + 6.

• the 12 region at the top: 6, 6
• the 10 region at the center: 5, 5
• the single-cell “>4” region at the bottom-right: 5
• the “=” region at the bottom: 1, 1, 1

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

Start with the “=” region at the left: all 4 cells must show the same number, and only 0 or 5 appear on enough dominoes to fill it.

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

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

Start with the 1 region at the top-right: the only way to make it with these dominoes is 0 + 0 + 1.

• the “=” region at the top-left: 0, 0, 0
• the 1 region at the top-right: 0, 0, 1
• the 11 region at the top-left: 5, 6
• the “=” region at the center: 2, 2
• the 10 region at the bottom-left: 4, 6
• the “=” region at the bottom-left: 3, 3
• the single-cell “>0” region at the right: 1
• the 12 region at the bottom-right: 6, 6
• the single-cell “>1” region at the bottom-left: 2

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