NYT Pips hints and answers, Saturday, August 29, 2026

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

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

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

Full solution (domino: first half → second half): 4–2: R1C4→R1C3; 6–6: R3C3→R3C4; 5–4: R3C1→R3C2; 5–6: R1C2→R1C1; 1–3: R2C3→R2C2.

Start with the region that must add up to 8 starting at row 2, column 2 (2 cells). It has the fewest ways to be satisfied.

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

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

Full solution (domino: first half → second half): 5–1: R2C2→R2C1; 6–3: R3C3→R3C2; 1–0: R3C1→R4C1; 6–6: R2C3→R2C4; 0–0: R1C5→R2C5; 4–2: R4C4→R3C4; 0–6: R4C2→R4C3.

Start with the region that must add up to 0 starting at row 1, column 5 (2 cells). It has the fewest ways to be satisfied.

Its cells hold 0, 0, in order from row 1, column 5.

Place the 0–1 domino with the 0 at row 1, column 5 and the 1 at row 1, column 4.

Full solution (domino: first half → second half): 1–3: R1C3→R1C2; 0–0: R6C1→R7C1; 3–2: R2C1→R2C2; 1–5: R7C4→R6C4; 6–6: R4C5→R5C5; 2–1: R4C2→R4C3; 0–4: R2C5→R3C5; 3–3: R7C2→R7C3; 6–0: R5C1→R4C1; 2–5: R7C5→R6C5; 0–1: R1C5→R1C4; 3–6: R6C2→R5C2; 1–1: R2C3→R3C3; 2–0: R3C2→R3C1.