NYT Pips hints and answers, Saturday, October 17, 2026

Start with the cell that must be greater than 5 at row 3, column 1. It has the fewest ways to be satisfied.

That cell is a 6, and only one domino has a 6: the 6–0.

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

Full solution (domino: first half → second half): 0–0: R2C4→R2C5; 6–0: R3C1→R3C2; 5–0: R3C3→R3C4; 0–1: R2C2→R1C2; 2–2: R1C3→R1C4.

Start with the cell that must equal 2 at row 2, column 2. It has the fewest ways to be satisfied.

That cell is a 2. Only 2 dominoes have a 2: 2–0, 6–2.

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

Full solution (domino: first half → second half): 2–0: R2C2→R2C3; 3–6: R4C1→R4C2; 5–5: R3C2→R3C3; 5–0: R3C1→R2C1; 3–3: R2C4→R3C4; 6–2: R4C3→R4C4; 4–6: R1C3→R1C4.

Start with the cell that must equal 5 at row 2, column 3. It has the fewest ways to be satisfied.

That cell is a 5. Only 3 dominoes have a 5: 3–5, 0–5, 6–5.

Place the 3–5 domino with the 3 at row 1, column 3 and the 5 at row 2, column 3.

Full solution (domino: first half → second half): 6–6: R5C8→R5C9; 3–5: R1C3→R2C3; 3–2: R3C3→R4C3; 1–4: R4C5→R5C5; 2–0: R1C6→R1C7; 6–3: R5C7→R6C7; 2–6: R2C5→R1C5; 1–6: R1C2→R1C1; 0–3: R3C6→R3C5; 0–0: R8C7→R9C7; 0–5: R5C1→R4C1; 4–3: R7C7→R7C8; 2–2: R2C1→R3C1; 6–5: R5C2→R5C3.