NYT Pips hints and answers, Tuesday, September 22, 2026

Start with the cell that must be greater than 5 at row 4, column 3. It has the fewest ways to be satisfied.

That cell is a 6, and only one domino has a 6: the 5–6.

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

Full solution (domino: first half → second half): 5–3: R3C1→R2C1; 5–6: R4C4→R4C3; 5–5: R4C1→R4C2; 2–1: R2C3→R1C3; 0–0: R1C1→R1C2.

Start with the region that must add up to 11 starting at row 2, column 3 (2 cells). It has the fewest ways to be satisfied.

Its cells hold 6, 5, in order from row 2, column 3.

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

Full solution (domino: first half → second half): 1–6: R4C2→R4C1; 6–6: R2C2→R2C3; 1–2: R3C2→R3C3; 0–5: R4C3→R4C4; 4–5: R1C4→R2C4; 3–5: R3C5→R4C5; 1–0: R3C1→R2C1.

Start with the cell that must equal 2 at row 5, column 7. It has the fewest ways to be satisfied.

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

Place the 1–2 domino with the 1 at row 5, column 6 and the 2 at row 5, column 7.

Full solution (domino: first half → second half): 1–3: R10C1→R9C1; 4–5: R10C7→R9C7; 3–2: R6C2→R6C3; 6–5: R7C7→R8C7; 1–2: R5C6→R5C7; 3–4: R2C1→R3C1; 3–3: R6C1→R7C1; 1–4: R5C5→R6C5; 0–6: R10C2→R10C3; 1–1: R3C4→R4C4; 3–6: R8C1→R8C2; 1–5: R2C4→R1C4; 6–4: R10C5→R10C6; 3–5: R7C5→R7C6; 4–2: R4C1→R4C2.