NYT Pips hints and answers, Friday, October 16, 2026

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

That cell is a 6, and only one domino has a 6: the 6–4.

Place the 6–4 domino with the 6 at row 5, column 3 and the 4 at row 5, column 2.

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

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

That cell is a 0. Only 3 dominoes have a 0: 0–3, 4–0, 1–0.

Place the 1–0 domino with the 1 at row 2, column 4 and the 0 at row 1, column 4.

Full solution (domino: first half → second half): 1–5: R3C1→R3C2; 5–4: R3C5→R2C5; 4–2: R2C2→R2C3; 5–5: R3C3→R3C4; 0–3: R4C5→R4C4; 4–0: R4C2→R4C1; 1–0: R2C4→R1C4.

Start with the cell that must equal 4 at row 1, column 2. It has the fewest ways to be satisfied.

That cell is a 4. Only 4 dominoes have a 4: 4–3, 4–5, 4–1, 2–4.

Place the 2–4 domino with the 2 at row 1, column 1 and the 4 at row 1, column 2.

Full solution (domino: first half → second half): 5–5: R5C7→R6C7; 4–3: R5C4→R6C4; 3–2: R1C7→R2C7; 0–1: R6C1→R5C1; 1–2: R5C2→R6C2; 4–5: R3C7→R4C7; 2–6: R3C2→R2C2; 1–5: R2C1→R3C1; 1–1: R2C5→R3C5; 0–0: R1C4→R2C4; 3–0: R4C4→R3C4; 4–1: R4C2→R4C1; 3–5: R3C6→R4C6; 2–4: R1C1→R1C2.