Pips answers for Thursday, January 1, 2026

Start with the 3 region at the center: only 0 + 0 + 3, or 0 + 1 + 2 can make it with these dominoes.

• the “=” region at the bottom-left: 5, 5
• the 3 region at the center: 0, 1, 2
• the single-cell “<2” region at the bottom: 1
• the single-cell 3 region at the bottom: 3

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

Start with the 10 region at the left: only 4 + 6, or 5 + 5 can make it with these dominoes.

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

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

Start with the “=” region at the left: all 4 cells must show the same number, and only 0 or 2 or 4 or 5 appear on enough dominoes to fill it.

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

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