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NYT Pips answers — Thursday, October 2, 2025
Pips #46 · Edited by Ian Livengood · Easy by Rodolfo Kurchan · Medium by Rodolfo Kurchan · Hard by Rodolfo Kurchan
Stuck on today's New York Times Pips? Below are three progressive hints for each difficulty, then the full solution as a tap-to-reveal board. Hints first — the answer only when you want it.
Easy Pips answer — October 2, 2025
4×4 board · 6 dominoes · 8 regions · by Rodolfo Kurchan
6 dominoes, 8 regions, a 4×4 grid — that's the easy puzzle for October 2, 2025. The mix is 3 sum regions, 1 "all equal", 1 "less than", 2 free regions. Best foothold — the single cell at row 2, column 4 is marked 2, so it can only hold a 2. After that, the single cell at row 4, column 2 is marked 1, so it can only hold a 1. Rack check: 1 double, 1 blank half, 2 sixes.
Hints
Hint 1 — where to start
The single cell at row 2, column 4 is marked 2, so it can only hold a 2. Find the domino with a 2 that fits there.
Hint 2 — first domino
Place the 2–6 domino with its 2 at row 2, column 4 and its 6 at row 3, column 4.
Hint 3 — second domino
Next, the 5–1 goes at row 3, column 2 → row 4, column 2. From there the rest should fall into place.
Solution
Tap any cell to reveal its pips, or reveal the whole board. Region rules: <2 must add up to less than 2; ≠ must all be different; = must all be the same number; 2 must add up to 2; 12 must add up to 12.
Solution as text
0 2 3 3 · 4 2 2 · 5 · 6 · 1 1 6
- 4–2 — 4 at row 2, column 2, 2 at row 2, column 3
- 2–0 — 2 at row 1, column 2, 0 at row 1, column 1
- 2–6 — 2 at row 2, column 4, 6 at row 3, column 4
- 5–1 — 5 at row 3, column 2, 1 at row 4, column 2
- 3–3 — 3 at row 1, column 3, 3 at row 1, column 4
- 1–6 — 1 at row 4, column 3, 6 at row 4, column 4
Medium Pips answer — October 2, 2025
6×5 board · 5 dominoes · 6 regions · by Rodolfo Kurchan
Medium today is a 6×5 board with 5 dominoes and 6 regions. Rules on the board: 3 sum regions, 1 "greater than", 1 free region. Best foothold — the single cell at row 1, column 5 is marked 1, so it can only hold a 1. After that, the pair at row 2, column 3 adds to 2; list the ways to make 2 with the rack you have.. Rack check: 2 blank halves.
Hints
Hint 1 — where to start
The single cell at row 1, column 5 is marked 1, so it can only hold a 1. Find the domino with a 1 that fits there.
Hint 2 — first domino
Place the 4–1 domino with its 4 at row 1, column 4 and its 1 at row 1, column 5.
Hint 3 — second domino
Next, the 2–0 goes at row 2, column 3 → row 1, column 3. From there the rest should fall into place.
Solution
Tap any cell to reveal its pips, or reveal the whole board. Region rules: >4 must add up to more than 4; ≠ must all be different; 1 must add up to 1; 2 must add up to 2; 4 must add up to 4.
Solution as text
5 3 0 4 1 · · 2 · · · · 0 · · · · 1 · · · · 3 · · · · 1 · ·
- 1–0 — 1 at row 4, column 3, 0 at row 3, column 3
- 1–3 — 1 at row 6, column 3, 3 at row 5, column 3
- 3–5 — 3 at row 1, column 2, 5 at row 1, column 1
- 4–1 — 4 at row 1, column 4, 1 at row 1, column 5
- 2–0 — 2 at row 2, column 3, 0 at row 1, column 3
Hard Pips answer — October 2, 2025
8×7 board · 14 dominoes · 17 regions · by Rodolfo Kurchan
14 dominoes, 17 regions, a 8×7 grid — that's the hard puzzle for October 2, 2025. You're working with 9 sum regions, 2 "all equal", 4 "less than", 1 "greater than", 1 free region. The single cell at row 2, column 5 is marked 0, so it can only hold a 0. After that, the single cell at row 4, column 6 is marked 1, so it can only hold a 1. Rack check: 4 doubles, 3 blank halves, 2 sixes.
Hints
Hint 1 — where to start
The single cell at row 2, column 5 is marked 0, so it can only hold a 0. Find the domino with a 0 that fits there.
Hint 2 — first domino
Place the 0–0 domino with its 0 at row 2, column 5 and its 0 at row 3, column 5.
Hint 3 — second domino
Next, the 1–1 goes at row 3, column 6 → row 4, column 6. From there the rest should fall into place.
Solution
Tap any cell to reveal its pips, or reveal the whole board. Region rules: >3 must add up to more than 3; = must all be the same number; 0 must add up to 0; 3 must add up to 3; 1 must add up to 1.
Solution as text
· · · 4 · · · · · 5 5 0 · · · 1 2 5 0 1 · · 2 1 5 4 1 · 4 3 0 4 6 6 1 · · · 2 · · · · · · 3 · · · · 2 3 3 3 1 ·
- 5–2 — 5 at row 2, column 3, 2 at row 3, column 3
- 2–1 — 2 at row 4, column 2, 1 at row 3, column 2
- 5–4 — 5 at row 2, column 4, 4 at row 1, column 4
- 1–1 — 1 at row 3, column 6, 1 at row 4, column 6
- 3–3 — 3 at row 7, column 4, 3 at row 8, column 4
- 4–6 — 4 at row 4, column 5, 6 at row 5, column 5
- 2–3 — 2 at row 8, column 2, 3 at row 8, column 3
- 3–4 — 3 at row 5, column 2, 4 at row 5, column 1
- 0–0 — 0 at row 2, column 5, 0 at row 3, column 5
- 6–1 — 6 at row 5, column 6, 1 at row 5, column 7
- 1–0 — 1 at row 4, column 3, 0 at row 5, column 3
- 4–2 — 4 at row 5, column 4, 2 at row 6, column 4
- 1–3 — 1 at row 8, column 6, 3 at row 8, column 5
- 5–5 — 5 at row 3, column 4, 5 at row 4, column 4
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