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NYT Pips answers — Sunday, November 2, 2025

Pips #77 · 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.

Jump to: Easy · Medium · Hard

Easy Pips answer — November 2, 2025

4×5 board · 5 dominoes · 6 regions · by Rodolfo Kurchan

5 dominoes, 6 regions, a 4×5 grid — that's the easy puzzle for November 2, 2025. You're working with 4 sum regions, 2 free regions. The way in: the single cell at row 2, column 5 is marked 2, so it can only hold a 2. After that, the two-cell region at row 1, column 3 adds to 12: both cells must be 6. Rack check: 1 double, 1 blank half, 3 sixes.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 4–6 goes at row 1, column 4 → 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: 12 must add up to 12; 10 must add up to 10; 2 must add up to 2.

12
10
2
2
Solution as text
  1. 4–6 — 4 at row 1, column 4, 6 at row 1, column 3
  2. 1–2 — 1 at row 2, column 4, 2 at row 2, column 5
  3. 6–0 — 6 at row 2, column 3, 0 at row 2, column 2
  4. 6–1 — 6 at row 4, column 4, 1 at row 3, column 4
  5. 5–5 — 5 at row 3, column 1, 5 at row 3, column 2

Medium Pips answer — November 2, 2025

5×4 board · 7 dominoes · 9 regions · by Rodolfo Kurchan

The medium board runs 5 rows by 4 columns: 7 dominoes to place across 9 regions. The mix is 4 sum regions, 1 "less than", 1 "greater than", 3 free regions. The single cell at row 4, column 1 is marked 2, so it can only hold a 2. After that, the two-cell region at row 3, column 2 adds to 11: the only split is 5 + 6.. Rack check: 1 double, 4 sixes.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 5–1 goes at row 3, column 2 → row 3, 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: 10 must add up to 10; <4 must add up to less than 4; >4 must add up to more than 4; 11 must add up to 11; 2 must add up to 2.

10
<4
>4
11
11
2
Solution as text
  1. 5–1 — 5 at row 3, column 2, 1 at row 3, column 3
  2. 1–6 — 1 at row 2, column 2, 6 at row 2, column 1
  3. 4–1 — 4 at row 1, column 3, 1 at row 2, column 3
  4. 6–4 — 6 at row 4, column 2, 4 at row 5, column 2
  5. 4–5 — 4 at row 5, column 4, 5 at row 4, column 4
  6. 6–6 — 6 at row 2, column 4, 6 at row 3, column 4
  7. 2–4 — 2 at row 4, column 1, 4 at row 3, column 1

Hard Pips answer — November 2, 2025

9×7 board · 14 dominoes · 14 regions · by Rodolfo Kurchan

The hard board runs 9 rows by 7 columns: 14 dominoes to place across 14 regions. The mix is 11 sum regions, 1 "all equal", 2 "greater than". The single cell at row 1, column 4 is marked 0, so it can only hold a 0. After that, the single cell at row 3, column 7 is marked 3, so it can only hold a 3. Rack check: 1 double, 5 blank halves, 4 sixes.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 3–2 goes at row 3, column 7 → row 4, column 7. 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; 8 must add up to 8; = must all be the same number; 3 must add up to 3; 6 must add up to 6.

0
8
=
3
6
8
0
0
10
2
12
8
>4
>4
Solution as text
  1. 6–5 — 6 at row 9, column 3, 5 at row 9, column 2
  2. 1–5 — 1 at row 5, column 5, 5 at row 6, column 5
  3. 5–0 — 5 at row 6, column 4, 0 at row 6, column 3
  4. 4–0 — 4 at row 1, column 5, 0 at row 1, column 4
  5. 6–4 — 6 at row 9, column 6, 4 at row 9, column 5
  6. 6–0 — 6 at row 8, column 3, 0 at row 7, column 3
  7. 1–0 — 1 at row 5, column 4, 0 at row 5, column 3
  8. 3–1 — 3 at row 4, column 2, 1 at row 4, column 3
  9. 1–1 — 1 at row 3, column 4, 1 at row 4, column 4
  10. 4–1 — 4 at row 2, column 5, 1 at row 2, column 4
  11. 1–6 — 1 at row 4, column 5, 6 at row 4, column 6
  12. 2–4 — 2 at row 7, column 5, 4 at row 8, column 5
  13. 3–2 — 3 at row 3, column 7, 2 at row 4, column 7
  14. 3–0 — 3 at row 4, column 1, 0 at row 5, column 1

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