NYT Pips hints and answers, Monday, October 12, 2026

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

That cell is a 3, and only one domino has a 3: the 3–2.

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

Full solution (domino: first half → second half): 2–4: R2C2→R2C3; 5–5: R1C3→R1C4; 2–0: R3C3→R3C4; 6–2: R3C1→R3C2; 3–2: R1C1→R1C2.

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. Only 2 dominoes have a 5: 5–4, 3–5.

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

Full solution (domino: first half → second half): 5–4: R3C6→R3C5; 2–1: R2C2→R2C1; 3–0: R3C4→R4C4; 0–4: R2C3→R2C4; 2–4: R3C2→R3C1; 3–5: R1C3→R1C4; 4–4: R2C5→R2C6; 3–2: R3C3→R4C3.

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

That cell is a 2. Only 5 dominoes have a 2: 0–2, 1–2, 2–3, 2–4, 2–5.

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

Full solution (domino: first half → second half): 0–1: R5C3→R5C4; 0–2: R3C6→R3C5; 0–3: R4C6→R5C6; 0–4: R4C3→R3C3; 0–5: R6C5→R5C5; 1–2: R4C2→R3C2; 1–3: R4C1→R5C1; 1–4: R2C1→R3C1; 1–5: R3C4→R2C4; 2–3: R1C4→R1C5; 2–4: R1C2→R2C2; 2–5: R6C4→R6C3; 3–4: R2C6→R2C5; 3–5: R6C2→R5C2; 4–5: R4C5→R4C4; 3–6: R1C3→R2C3.