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NYT Pips answers — Monday, December 22, 2025
Pips #127 · Edited by Ian Livengood · Easy by Ian Livengood · Medium by Ian Livengood · 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 — December 22, 2025
4×5 board · 5 dominoes · 8 regions · by Ian Livengood
Easy today is a 4×5 board with 5 dominoes and 8 regions. Rules on the board: 4 sum regions, 1 "greater than", 3 free regions. The way in: the single cell at row 2, column 4 is marked 4, so it can only hold a 4. After that, the single cell at row 3, column 1 is marked 1, so it can only hold a 1. Rack check: 1 blank half, 3 sixes.
Hints
Hint 1 — where to start
The single cell at row 2, column 4 is marked 4, so it can only hold a 4. Find the domino with a 4 that fits there.
Hint 2 — first domino
Place the 4–5 domino with its 4 at row 2, column 4 and its 5 at row 2, column 5.
Hint 3 — second domino
Next, the 2–1 goes at row 3, column 2 → row 3, column 1. From there the rest should fall into place.
Solution
Tap any cell to reveal its pips, or reveal the whole board. Region rules: 6 must add up to 6; 7 must add up to 7; 4 must add up to 4; 1 must add up to 1; >1 must add up to more than 1.
Solution as text
2 6 0 · · · · 6 4 5 1 2 1 · · · · 6 · ·
- 1–6 — 1 at row 3, column 3, 6 at row 4, column 3
- 2–1 — 2 at row 3, column 2, 1 at row 3, column 1
- 0–6 — 0 at row 1, column 3, 6 at row 2, column 3
- 4–5 — 4 at row 2, column 4, 5 at row 2, column 5
- 2–6 — 2 at row 1, column 1, 6 at row 1, column 2
Medium Pips answer — December 22, 2025
5×5 board · 7 dominoes · 8 regions · by Ian Livengood
7 dominoes, 8 regions, a 5×5 grid — that's the medium puzzle for December 22, 2025. The mix is 3 sum regions, 1 "all equal", 3 "greater than". The way in: the single cell at row 2, column 1 is marked 4, so it can only hold a 4. After that, the single cell at row 5, column 2 is marked 1, so it can only hold a 1. Rack check: 1 double, 1 blank half, 1 six.
Hints
Hint 1 — where to start
The single cell at row 2, column 1 is marked 4, so it can only hold a 4. Find the domino with a 4 that fits there.
Hint 2 — first domino
Place the 4–3 domino with its 4 at row 2, column 1 and its 3 at row 3, column 1.
Hint 3 — second domino
Next, the 5–1 goes at row 5, column 3 → row 5, 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: >5 must add up to more than 5; ≠ must all be different; 4 must add up to 4; >3 must add up to more than 3; = must all be the same number.
Solution as text
· 6 4 · · 4 · 0 5 · 3 3 1 5 5 · · 2 4 · · 1 5 · ·
- 1–3 — 1 at row 3, column 3, 3 at row 3, column 2
- 5–5 — 5 at row 3, column 4, 5 at row 3, column 5
- 4–3 — 4 at row 2, column 1, 3 at row 3, column 1
- 5–0 — 5 at row 2, column 4, 0 at row 2, column 3
- 5–1 — 5 at row 5, column 3, 1 at row 5, column 2
- 4–6 — 4 at row 1, column 3, 6 at row 1, column 2
- 4–2 — 4 at row 4, column 4, 2 at row 4, column 3
Hard Pips answer — December 22, 2025
6×6 board · 16 dominoes · 14 regions · by Rodolfo Kurchan
16 dominoes, 14 regions, a 6×6 grid — that's the hard puzzle for December 22, 2025. You're working with 8 sum regions, 2 "all equal", 2 "less than", 2 "greater than". The single cell at row 1, column 3 is marked 4, so it can only hold a 4. After that, the single cell at row 1, column 4 is marked 0, so it can only hold a 0. Rack check: 5 doubles, 6 blank halves, 7 sixes.
Hints
Hint 1 — where to start
The single cell at row 1, column 3 is marked 4, so it can only hold a 4. Find the domino with a 4 that fits there.
Hint 2 — first domino
Place the 2–4 domino with its 2 at row 2, column 3 and its 4 at row 1, column 3.
Hint 3 — second domino
Next, the 0–5 goes at row 1, column 4 → row 2, column 4. From there the rest should fall into place.
Solution
Tap any cell to reveal its pips, or reveal the whole board. Region rules: 0 must add up to 0; <4 must add up to less than 4; 4 must add up to 4; 12 must add up to 12; = must all be the same number.
Solution as text
0 1 4 0 6 0 0 · 2 5 6 3 0 · 2 5 1 1 0 2 2 6 1 1 · 6 6 6 2 · 3 3 3 6 2 5
- 1–1 — 1 at row 4, column 5, 1 at row 4, column 6
- 2–6 — 2 at row 5, column 5, 6 at row 5, column 4
- 2–2 — 2 at row 3, column 3, 2 at row 4, column 3
- 6–3 — 6 at row 6, column 4, 3 at row 6, column 3
- 0–0 — 0 at row 2, column 1, 0 at row 3, column 1
- 5–6 — 5 at row 3, column 4, 6 at row 4, column 4
- 2–4 — 2 at row 2, column 3, 4 at row 1, column 3
- 6–0 — 6 at row 1, column 5, 0 at row 1, column 6
- 3–3 — 3 at row 6, column 1, 3 at row 6, column 2
- 6–6 — 6 at row 5, column 2, 6 at row 5, column 3
- 1–0 — 1 at row 1, column 2, 0 at row 1, column 1
- 1–6 — 1 at row 3, column 5, 6 at row 2, column 5
- 2–5 — 2 at row 6, column 5, 5 at row 6, column 6
- 1–3 — 1 at row 3, column 6, 3 at row 2, column 6
- 0–2 — 0 at row 4, column 1, 2 at row 4, column 2
- 0–5 — 0 at row 1, column 4, 5 at row 2, column 4
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