NYT Pips hints and answers, Wednesday, September 16, 2026

Start with the cell that must be less than 1 at row 6, column 1. It has the fewest ways to be satisfied.

That cell is a 0. Only 2 dominoes have a 0: 0–1, 3–0.

Place the 3–0 domino with the 3 at row 6, column 2 and the 0 at row 6, column 1.

Full solution (domino: first half → second half): 0–1: R2C4→R1C4; 4–4: R4C2→R4C3; 5–3: R6C4→R6C3; 3–0: R6C2→R6C1; 6–1: R2C1→R1C1.

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

That cell is a 3, and only one domino has a 3: the 3–1.

Place the 3–1 domino with the 3 at row 3, column 1 and the 1 at row 3, column 2.

Full solution (domino: first half → second half): 4–1: R4C3→R4C4; 3–1: R3C1→R3C2; 5–5: R3C3→R3C4; 6–1: R1C2→R2C2; 5–1: R2C4→R2C3; 4–4: R4C1→R4C2; 1–0: R4C5→R3C5.

Start with the cell that must equal 4 at row 1, column 7. It has the fewest ways to be satisfied.

That cell is a 4. Only 4 dominoes have a 4: 2–4, 5–4, 4–1, 4–4.

Place the 5–4 domino with the 5 at row 1, column 8 and the 4 at row 1, column 7.

Full solution (domino: first half → second half): 2–4: R8C2→R8C1; 5–6: R2C8→R3C8; 2–2: R1C6→R2C6; 5–4: R1C8→R1C7; 3–2: R1C3→R2C3; 5–5: R5C2→R5C3; 4–1: R5C5→R5C6; 5–2: R6C2→R7C2; 4–4: R7C7→R8C7; 1–6: R5C7→R6C7; 5–3: R3C7→R3C6; 0–0: R5C8→R5C9; 5–1: R3C4→R3C3.