Pips answers for Tuesday, December 2, 2025

Start with the “=” region at the right: all 3 cells must show the same number, and only 6s appear on enough dominoes to fill it.

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

1. 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 1–6 domino horizontally on row 4, column 1 and row 4, column 2.
2. The single-cell “<2” region at the top-right must total less than 2 — nothing else fits. Place the 1–2 domino horizontally on row 1, column 1 and row 1, column 2.
3. The “<6” region at the top-left must total less than 6 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–4 domino vertically on row 2, column 1 and row 3, column 1.
4. Row 2, column 2 has only one open neighbor, so its domino must run to row 3, column 2; the “=” region at the right needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 6–6 domino vertically on row 2, column 2 and row 3, column 2.

Start with the “=” region at the right: all 3 cells must show the same number, and only 4s appear on enough dominoes to fill it.

• the single-cell 3 region at the top-right: 3
• the 0 region at the center: 0, 0
• the single-cell “<5” region at the right: 4
• the “=” region at the right: 4, 4, 4
• the 17 region at the center: 5, 6, 6

1. The 17 region at the center needs a total of exactly 17 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 6–6 domino horizontally on row 3, column 3 and row 3, column 4.
2. The “=” region at the right needs every pip to match — nothing else fits. Place the 4–4 domino vertically on row 2, column 5 and row 3, column 5.
3. The 17 region at the center needs a total of exactly 17 (2 of its 3 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 1–5 domino vertically on row 3, column 2 and row 4, column 2.
4. Row 4, column 5 has only one open neighbor, so its domino must run to row 4, column 4; the “=” region at the right needs every pip to match (2 of its 3 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 2–4 domino horizontally on row 4, column 4 and row 4, column 5.
5. The 0 region at the center needs a total of exactly 0 — 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 2, column 1 and row 2, column 2.
6. The 0 region at the center needs a total of exactly 0 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–5 domino vertically on row 1, column 3 and row 2, column 3.
7. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the single-cell 3 region at the top-right needs a total of exactly 3 — nothing else fits. Place the 3–4 domino vertically on row 1, column 4 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-left: the only way to make it with these dominoes is 0 + 0 + 0 + 0.

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

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