Pips answers for Tuesday, March 17, 2026

Start with the 3 region at the left: the only way to make it with these dominoes is 1 + 2.

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

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

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

1. The “<3” region at the right must total less than 3 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–5 domino vertically on row 4, column 4 and row 5, column 4.
2. The “<3” region at the right must total less than 3 (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 0–3 domino horizontally on row 3, column 4 and row 3, column 5.
3. Row 5, column 3 has only one open neighbor, so its domino must run to row 5, column 2; the 11 region at the bottom needs a total of exactly 11 (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 0–6 domino horizontally on row 5, column 2 and row 5, column 3.
4. The “≠” region at the top needs every pip to be different — 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 2, column 2 and row 2, column 3.
5. Row 2, column 4 has only one open neighbor, so its domino must run to row 1, column 4; the “≠” region at the top needs every pip to be different (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 3–6 domino vertically on row 1, column 4 and row 2, column 4.
6. The “=” region at the left 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 1–1 domino vertically on row 3, column 2 and row 4, column 2.
7. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; the 11 region at the left needs a total of exactly 11 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–6 domino vertically on row 3, column 1 and row 4, column 1.

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

Start with the 18 region at the left: the only way to make it with these dominoes is 6 + 6 + 6.

• the 6 region at the top: 2, 2, 2
• the 3 region at the top-right: 1, 1, 1
• the single-cell “<2” region at the top-left: 0
• the “=” region at the top: 5, 5, 5, 5, 5
• the single-cell 6 region at the top-right: 6
• the single-cell “>2” region at the left: 3
• the 18 region at the left: 6, 6, 6
• the single-cell 2 region at the center: 2
• the 6 region at the center: 3, 3
• the single-cell “<2” region at the left: 0
• the 10 region at the left: 5, 5
• the 8 region at the center: 4, 4
• the single-cell 2 region at the bottom-left: 2
• the single-cell 1 region at the bottom: 1
• the “>9” region at the bottom: 6, 6

1. The 3 region at the top-right needs a total of exactly 3 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–1 domino horizontally on row 2, column 8 and row 2, column 9.
2. The single-cell 1 region at the bottom needs a total of exactly 1 — 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 6, column 4 and row 7, column 4.
3. The 3 region at the top-right needs a total of exactly 3 (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 1–2 domino horizontally on row 2, column 6 and row 2, column 7.
4. Row 1, column 6 has only one open neighbor, so its domino must run to row 1, column 7; the 6 region at the top needs a total of exactly 6 (1 of its 3 cells is 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 1, column 6 and row 1, column 7.
5. The 10 region at the left 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 2–5 domino horizontally on row 6, column 2 and row 6, column 3.
6. The 18 region at the left needs a total of exactly 18 — 3 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 5, column 1 and row 5, column 2.
7. The single-cell “<2” region at the top-left must total less than 2 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–5 domino horizontally on row 3, column 2 and row 3, column 3.
8. The “=” region at the top needs every pip to match (1 of its 5 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 3, column 4 and row 3, column 5.
9. The “=” region at the top needs every pip to match (3 of its 5 cells are already filled) — nothing else fits. Place the 5–6 domino horizontally on row 3, column 6 and row 3, column 7.
10. The 18 region at the left needs a total of exactly 18 (1 of its 3 cells is already filled) — nothing else fits. Place the 2–6 domino horizontally on row 4, column 3 and row 4, column 4.
11. Row 8, column 4 has only one open neighbor, so its domino must run to row 8, column 5; the “>9” region at the bottom must total more than 9 — nothing else fits. Place the 6–6 domino horizontally on row 8, column 4 and row 8, column 5.
12. Row 4, column 1 has only one open neighbor, so its domino must run to row 4, column 2; the single-cell “>2” region at the left must total more than 2 — nothing else fits. Place the 3–6 domino horizontally on row 4, column 1 and row 4, column 2.
13. 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 3–5 domino horizontally on row 4, column 5 and row 4, column 6.
14. Row 5, column 3 has only one open neighbor, so its domino must run to row 5, column 4; 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 4–5 domino horizontally on row 5, column 3 and row 5, column 4.
15. The 8 region at the center needs a total of exactly 8 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–4 domino vertically on row 5, column 5 and row 6, column 5.
16. Row 5, column 6 has only one open neighbor, so its domino must run to row 6, column 6; the 6 region at the center needs a total of exactly 6 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–3 domino vertically on row 5, column 6 and row 6, column 6.