Pips answers for Friday, March 6, 2026

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

• the “=” region at the top-right: 2, 2
• the “=” region at the center: 4, 4, 4, 4, 4
• the single-cell “<2” region at the left: 0
• the single-cell “>4” region at the right: 6

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

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

• the “=” region at the top-right: 4, 4
• the “=” region at the left: 3, 3, 3
• the 8 region at the center: 2, 6
• the single-cell “<6” region at the left: 0
• the “=” region at the bottom: 5, 5
• the “=” region at the bottom-right: 1, 1

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

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

• the 15 region at the top-left: 4, 5, 6
• the “=” region at the top: 3, 3, 3
• the 1 region at the top-right: 0, 0, 1
• the single-cell 3 region at the top: 3
• the “=” region at the left: 0, 0, 0
• the 12 region at the center: 6, 6
• the single-cell “>3” region at the left: 4
• the single-cell 4 region at the bottom-left: 4
• the single-cell “>3” region at the bottom: 5
• the 4 region at the bottom-left: 1, 3
• the 7 region at the bottom-right: 1, 1, 5
• the single-cell “<4” region at the bottom: 3

1. The 1 region at the top-right needs a total of exactly 1 — nothing else fits. Place the 0–0 domino vertically on row 3, column 7 and row 4, column 7.
2. The 4 region at the bottom-left needs a total of exactly 4 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–4 domino vertically on row 7, column 2 and row 8, column 2.
3. Row 9, column 2 has only one open neighbor, so its domino must run to row 9, column 3; the 4 region at the bottom-left needs a total of exactly 4 (1 of its 2 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–3 domino horizontally on row 9, column 2 and row 9, column 3.
4. Row 2, column 7 has only one open neighbor, so its domino must run to row 1, column 7; 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; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino vertically on row 1, column 7 and row 2, column 7.
5. Row 1, column 5 has only one open neighbor, so its domino must run to row 2, column 5; the “=” region at the top needs every pip to match — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–3 domino vertically on row 1, column 5 and row 2, column 5.
6. Row 3, column 5 has only one open neighbor, so its domino must run to row 4, column 5; the “=” region at the top needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 3–6 domino vertically on row 3, column 5 and row 4, column 5.
7. Row 3, column 3 has only one open neighbor, so its domino must run to row 4, column 3; the single-cell 3 region at the top needs a total of exactly 3 — nothing else fits. Place the 0–3 domino vertically on row 3, column 3 and row 4, column 3.
8. The 15 region at the top-left needs a total of exactly 15 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–6 domino vertically on row 3, column 1 and row 4, column 1.
9. Row 4, column 2 has only one open neighbor, so its domino must run to row 5, column 2; the “=” region at the left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 0–4 domino vertically on row 4, column 2 and row 5, column 2.
10. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the 12 region at the center needs a total of exactly 12 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–6 domino vertically on row 4, column 4 and row 5, column 4.
11. The 7 region at the bottom-right needs a total of exactly 7 — nothing else fits. Place the 1–1 domino horizontally on row 9, column 5 and row 9, column 6.
12. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 15 region at the top-left needs a total of exactly 15 (1 of its 3 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–5 domino vertically on row 1, column 1 and row 2, column 1.
13. Row 7, column 5 has only one open neighbor, so its domino must run to row 8, column 5; the single-cell “>3” region at the bottom must total more than 3 — nothing else fits. Place the 5–5 domino vertically on row 7, column 5 and row 8, column 5.

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