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NYT Pips answers — Saturday, November 1, 2025

Pips #76 · 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 1, 2025

3×5 board · 4 dominoes · 5 regions · by Rodolfo Kurchan

4 dominoes, 5 regions, a 3×5 grid — that's the easy puzzle for November 1, 2025. You're working with 1 sum region, 2 "all equal", 2 free regions. The way in: the single cell at row 3, column 3 is marked 4, so it can only hold a 4. After that, the "=" region starting at row 1, column 1 has 3 cells that must all match — every 3 in the puzzle should come here.. Rack check: 1 double, 1 blank half, 2 sixes.

Hints

Hint 1 — where to start

The single cell at row 3, 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 0–4 domino with its 0 at row 2, column 3 and its 4 at row 3, column 3.

Hint 3 — second domino

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

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

Medium Pips answer — November 1, 2025

4×6 board · 8 dominoes · 10 regions · by Rodolfo Kurchan

The medium board runs 4 rows by 6 columns: 8 dominoes to place across 10 regions. The mix is 6 sum regions, 1 "all equal", 3 free regions. The single cell at row 2, column 5 is marked 6, so it can only hold a 6. After that, the single cell at row 3, column 3 is marked 0, so it can only hold a 0. Rack check: 3 doubles, 1 blank half, 4 sixes.

Hints

Hint 1 — where to start

The single cell at row 2, column 5 is marked 6, so it can only hold a 6. Find the domino with a 6 that fits there.

Hint 2 — first domino

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

Hint 3 — second domino

Next, the 3–0 goes at row 4, column 3 → 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: = must all be the same number; 12 must add up to 12; 5 must add up to 5; 6 must add up to 6; 0 must add up to 0.

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

Hard Pips answer — November 1, 2025

5×8 board · 14 dominoes · 11 regions · by Rodolfo Kurchan

Hard today is a 5×8 board with 14 dominoes and 11 regions. You're working with 5 sum regions, 5 "all equal", 1 free region. The single cell at row 4, column 5 is marked 3, so it can only hold a 3. After that, the single cell at row 5, column 2 is marked 1, so it can only hold a 1. Rack check: 6 doubles, 3 blank halves, 4 sixes.

Hints

Hint 1 — where to start

The single cell at row 4, column 5 is marked 3, so it can only hold a 3. Find the domino with a 3 that fits there.

Hint 2 — first domino

Place the 4–3 domino with its 4 at row 5, column 5 and its 3 at row 4, column 5.

Hint 3 — second domino

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

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

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