NYT Pips hints and answers, Thursday, September 10, 2026

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

That cell is a 3, and only one domino has a 3: the 0–3.

Place the 0–3 domino with the 0 at row 3, column 4 and the 3 at row 3, column 3.

Full solution (domino: first half → second half): 0–4: R2C4→R1C4; 6–5: R2C2→R2C1; 4–6: R1C3→R2C3; 1–6: R1C1→R1C2; 0–3: R3C4→R3C3.

Start with the cell that must equal 5 at row 2, column 4. It has the fewest ways to be satisfied.

That cell is a 5, and only one domino has a 5: the 0–5.

Place the 0–5 domino with the 0 at row 2, column 3 and the 5 at row 2, column 4.

Full solution (domino: first half → second half): 4–6: R3C2→R3C1; 0–0: R3C4→R4C4; 3–3: R4C2→R4C3; 0–1: R1C5→R2C5; 2–2: R1C3→R1C4; 0–5: R2C3→R2C4; 6–0: R2C1→R2C2.

Start with the cell that must equal 0 at row 1, column 2. It has the fewest ways to be satisfied.

That cell is a 0. Only 6 dominoes have a 0: 0–0, 3–0, 0–1, 0–6, 2–0, 5–0.

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

Full solution (domino: first half → second half): 1–3: R7C5→R7C6; 5–4: R2C3→R3C3; 1–5: R9C9→R8C9; 0–0: R4C7→R5C7; 1–2: R7C1→R8C1; 3–0: R3C6→R3C5; 6–2: R2C7→R1C7; 0–1: R7C3→R7C2; 2–2: R9C1→R9C2; 0–6: R1C2→R1C3; 5–5: R7C8→R7C9; 2–0: R9C3→R8C3; 4–4: R9C5→R9C6; 5–0: R3C9→R3C8.