Pips answers for Friday, March 20, 2026

Start with the 10 region at the right: the only way to make it with these dominoes is 0 + 5 + 5.

• the 10 region at the right: 0, 5, 5
• the “=” region at the left: 3, 3, 3
• the single-cell “>5” region at the bottom-left: 6

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

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

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

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

Start with the 6 region at the top-right — it has the fewest pip combinations on the board, so the dominoes around it are the easiest to pin down.

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

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

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