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NYT Pips answers — Sunday, November 16, 2025

Pips #91 · 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 — November 16, 2025

4×5 board · 5 dominoes · 7 regions · by Ian Livengood

Easy today is a 4×5 board with 5 dominoes and 7 regions. Rules on the board: 2 sum regions, 2 "all equal", 3 free regions. The way in: the single cell at row 1, column 4 is marked 2, so it can only hold a 2. After that, the single cell at row 4, column 2 is marked 2, so it can only hold a 2. Rack check: 1 double, 2 blank halves.

Hints

Hint 1 — where to start

The single cell at row 1, 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 2–0 domino with its 2 at row 1, column 4 and its 0 at row 1, column 5.

Hint 3 — second domino

Next, the 2–5 goes at row 4, column 2 → row 3, 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 2; = must all be the same number.

2
=
=
2
Solution as text
  1. 1–3 — 1 at row 2, column 3, 3 at row 3, column 3
  2. 0–4 — 0 at row 2, column 5, 4 at row 2, column 4
  3. 1–1 — 1 at row 2, column 1, 1 at row 2, column 2
  4. 2–5 — 2 at row 4, column 2, 5 at row 3, column 2
  5. 2–0 — 2 at row 1, column 4, 0 at row 1, column 5

Medium Pips answer — November 16, 2025

5×6 board · 8 dominoes · 13 regions · by Rodolfo Kurchan

The medium board runs 5 rows by 6 columns: 8 dominoes to place across 13 regions. The mix is 7 sum regions, 6 free regions. The single cell at row 2, column 3 is marked 4, so it can only hold a 4. After that, the single cell at row 4, column 5 is marked 2, so it can only hold a 2. Rack check: 2 doubles, 5 blank halves, 3 sixes.

Hints

Hint 1 — where to start

The single cell at row 2, 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 6–4 domino with its 6 at row 2, column 2 and its 4 at row 2, column 3.

Hint 3 — second domino

Next, the 2–0 goes at row 4, column 5 → row 3, column 5. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: 7 must add up to 7; 4 must add up to 4; 8 must add up to 8; 6 must add up to 6; 2 must add up to 2.

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

Hard Pips answer — November 16, 2025

7×4 board · 12 dominoes · 12 regions · by Rodolfo Kurchan

12 dominoes, 12 regions, a 7×4 grid — that's the hard puzzle for November 16, 2025. You're working with 7 sum regions, 4 "all equal", 1 free region. The way in: the single cell at row 3, column 4 is marked 1, so it can only hold a 1. After that, the single cell at row 4, column 2 is marked 3, so it can only hold a 3. Rack check: 2 doubles, 3 blank halves, 3 sixes.

Hints

Hint 1 — where to start

The single cell at row 3, column 4 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 5–1 domino with its 5 at row 2, column 4 and its 1 at row 3, column 4.

Hint 3 — second domino

Next, the 3–5 goes at row 4, column 2 → row 4, 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: 12 must add up to 12; 8 must add up to 8; 10 must add up to 10; = must all be the same number; 1 must add up to 1.

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

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