NYT Pips hints and answers, Wednesday, September 9, 2026

Start with the cell that must be greater than 4 at row 1, column 4. It has the fewest ways to be satisfied.

That cell is a 5, and only one domino has a 5: the 1–5.

Place the 1–5 domino with the 1 at row 1, column 3 and the 5 at row 1, column 4.

Full solution (domino: first half → second half): 1–1: R2C3→R2C4; 3–2: R1C1→R1C2; 1–0: R3C3→R3C2; 0–6: R3C1→R2C1; 1–5: R1C3→R1C4.

Start with the cell that must equal 0 at row 3, column 5. It has the fewest ways to be satisfied.

That cell is a 0, and only one domino has a 0: the 0–3.

Place the 0–3 domino with the 0 at row 3, column 5 and the 3 at row 3, column 4.

Full solution (domino: first half → second half): 3–1: R4C2→R3C2; 5–5: R4C1→R5C1; 2–3: R1C4→R2C4; 5–6: R3C3→R2C3; 0–3: R3C5→R3C4; 5–3: R5C3→R5C2; 5–4: R4C3→R4C4.

Start with the cell that must equal 3 at row 1, column 9. It has the fewest ways to be satisfied.

That cell is a 3. Only 3 dominoes have a 3: 3–1, 3–3, 3–4.

Place the 3–1 domino with the 3 at row 1, column 9 and the 1 at row 1, column 8.

Full solution (domino: first half → second half): 6–2: R1C2→R1C1; 5–0: R1C3→R2C3; 1–1: R1C5→R1C6; 0–0: R3C2→R3C3; 4–4: R7C5→R8C5; 3–1: R1C9→R1C8; 0–6: R3C7→R4C7; 4–2: R5C3→R4C3; 0–1: R2C7→R1C7; 5–6: R5C7→R6C7; 2–2: R5C5→R6C5; 3–3: R2C1→R3C1; 1–5: R9C6→R9C5; 3–4: R4C1→R5C1.