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NYT Pips answers — Sunday, December 28, 2025

Pips #133 · Edited by Ian Livengood · Easy by Ian Livengood · 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.

Jump to: Easy · Medium · Hard

Easy Pips answer — December 28, 2025

5×3 board · 5 dominoes · 6 regions · by Ian Livengood

5 dominoes, 6 regions, a 5×3 grid — that's the easy puzzle for December 28, 2025. You're working with 3 sum regions, 3 "all equal". The single cell at row 1, column 2 is marked 5, so it can only hold a 5. After that, the single cell at row 4, column 1 is marked 5, so it can only hold a 5. Rack check: 1 double, 2 sixes.

Hints

Hint 1 — where to start

The single cell at row 1, column 2 is marked 5, so it can only hold a 5. Find the domino with a 5 that fits there.

Hint 2 — first domino

Place the 5–6 domino with its 5 at row 1, column 2 and its 6 at row 1, column 3.

Hint 3 — second domino

Next, the 5–3 goes at row 4, column 1 → 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: 5 must add up to 5; = must all be the same number; 4 must add up to 4.

5
=
=
5
=
4
Solution as text
  1. 5–6 — 5 at row 1, column 2, 6 at row 1, column 3
  2. 5–3 — 5 at row 4, column 1, 3 at row 4, column 2
  3. 2–2 — 2 at row 5, column 1, 2 at row 5, column 2
  4. 3–4 — 3 at row 4, column 3, 4 at row 5, column 3
  5. 6–3 — 6 at row 2, column 3, 3 at row 3, column 3

Medium Pips answer — December 28, 2025

4×5 board · 7 dominoes · 7 regions · by Rodolfo Kurchan

Medium today is a 4×5 board with 7 dominoes and 7 regions. Rules on the board: 3 sum regions, 1 "greater than", 2 free regions. Best foothold — the single cell at row 3, column 1 is marked 0, so it can only hold a 0. After that, the pair at row 4, column 2 adds to 6; list the ways to make 6 with the rack you have.. Rack check: 2 doubles, 2 blank halves, 1 six.

Hints

Hint 1 — where to start

The single cell at row 3, column 1 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 3–0 domino with its 3 at row 3, column 2 and its 0 at row 3, column 1.

Hint 3 — second domino

Next, the 3–1 goes at row 4, column 2 → row 4, 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: >3 must add up to more than 3; ≠ must all be different; 8 must add up to 8; 0 must add up to 0; 6 must add up to 6.

>3
≠
8
0
6
Solution as text
  1. 3–0 — 3 at row 3, column 2, 0 at row 3, column 1
  2. 6–2 — 6 at row 1, column 3, 2 at row 2, column 3
  3. 1–1 — 1 at row 2, column 1, 1 at row 2, column 2
  4. 3–2 — 3 at row 4, column 3, 2 at row 4, column 4
  5. 2–0 — 2 at row 3, column 4, 0 at row 3, column 3
  6. 3–1 — 3 at row 4, column 2, 1 at row 4, column 1
  7. 2–2 — 2 at row 2, column 4, 2 at row 2, column 5

Hard Pips answer — December 28, 2025

8×4 board · 12 dominoes · 11 regions · by Rodolfo Kurchan

Hard today is a 8×4 board with 12 dominoes and 11 regions. Rules on the board: 3 sum regions, 3 "all equal", 2 "greater than", 3 free regions. The way in: the "=" region starting at row 1, column 4 has 3 cells that must all match — every 1 in the puzzle should come here.. After that, the "=" region starting at row 4, column 1 has 3 cells that must all match — every 0 in the puzzle should come here.. Rack check: 3 doubles, 3 blank halves, 3 sixes.

Hints

Hint 1 — where to start

The "=" region starting at row 1, column 4 has 3 cells that must all match — every 1 in the puzzle should come here.

Hint 2 — first domino

Place the 1–1 domino with its 1 at row 1, column 4 and its 1 at row 2, column 4.

Hint 3 — second domino

Next, the 0–4 goes at row 4, column 1 → 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: 15 must add up to 15; >4 must add up to more than 4; = must all be the same number; >3 must add up to more than 3.

15
>4
=
15
=
15
>3
=
Solution as text
  1. 2–0 — 2 at row 7, column 1, 0 at row 6, column 1
  2. 2–6 — 2 at row 5, column 4, 6 at row 5, column 3
  3. 0–4 — 0 at row 4, column 1, 4 at row 3, column 1
  4. 6–5 — 6 at row 2, column 1, 5 at row 1, column 1
  5. 5–2 — 5 at row 3, column 2, 2 at row 4, column 2
  6. 1–1 — 1 at row 1, column 4, 1 at row 2, column 4
  7. 4–5 — 4 at row 1, column 2, 5 at row 1, column 3
  8. 4–4 — 4 at row 6, column 3, 4 at row 6, column 4
  9. 0–5 — 0 at row 5, column 1, 5 at row 5, column 2
  10. 2–2 — 2 at row 8, column 1, 2 at row 8, column 2
  11. 4–2 — 4 at row 6, column 2, 2 at row 7, column 2
  12. 1–6 — 1 at row 2, column 3, 6 at row 2, column 2

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