NYT Pips hints and answers, Tuesday, September 29, 2026

Start with the cell that must be less than 2 at row 1, column 3. It has the fewest ways to be satisfied.

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

Place the 5–0 domino with the 5 at row 2, column 3 and the 0 at row 1, column 3.

Full solution (domino: first half → second half): 5–0: R2C3→R1C3; 2–5: R4C2→R4C3; 4–3: R5C2→R5C1; 0–2: R2C2→R3C2; 3–0: R4C1→R3C1.

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

That cell is a 1, and only one domino has a 1: the 1–2.

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

Full solution (domino: first half → second half): 4–5: R1C2→R1C3; 2–6: R3C3→R2C3; 1–2: R3C1→R3C2; 4–4: R2C1→R2C2; 6–6: R2C5→R3C5; 3–3: R3C4→R4C4; 0–5: R1C4→R2C4.

Start with the cell that must equal 4 at row 1, column 5. It has the fewest ways to be satisfied.

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

Place the 3–4 domino with the 3 at row 2, column 5 and the 4 at row 1, column 5.

Full solution (domino: first half → second half): 6–3: R3C1→R3C2; 3–3: R1C3→R1C4; 3–2: R6C1→R7C1; 1–1: R5C3→R6C3; 5–5: R3C5→R4C5; 5–3: R3C4→R2C4; 2–5: R7C4→R6C4; 1–6: R5C4→R5C5; 0–3: R2C2→R2C3; 4–4: R6C5→R7C5; 3–4: R2C5→R1C5; 4–2: R7C2→R7C3; 1–4: R5C1→R4C1; 2–0: R1C1→R2C1.