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NYT Pips answers — Friday, November 14, 2025

Pips #89 · 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.

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Easy Pips answer — November 14, 2025

3×4 board · 4 dominoes · 4 regions · by Ian Livengood

4 dominoes, 4 regions, a 3×4 grid — that's the easy puzzle for November 14, 2025. You're working with 2 sum regions, 2 "all equal". The way in: the single cell at row 1, column 4 is marked 5, so it can only hold a 5. After that, the single cell at row 3, column 3 is marked 1, so it can only hold a 1. Rack check: 1 double, 4 sixes.

Hints

Hint 1 — where to start

The single cell at row 1, column 4 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 6–5 domino with its 6 at row 2, column 4 and its 5 at row 1, column 4.

Hint 3 — second domino

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

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

Medium Pips answer — November 14, 2025

3×6 board · 7 dominoes · 7 regions · by Rodolfo Kurchan

7 dominoes, 7 regions, a 3×6 grid — that's the medium puzzle for November 14, 2025. You're working with 2 sum regions, 3 "all equal", 1 "greater than", 1 free region. The "=" region starting at row 1, column 5 has 3 cells that must all match — every 1 in the puzzle should come here.. After that, the "=" region starting at row 2, column 2 has 3 cells that must all match — every 2 in the puzzle should come here.. Rack check: 1 double, 2 blank halves, 1 six.

Hints

Hint 1 — where to start

The "=" region starting at row 1, column 5 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 5 and its 1 at row 2, column 5.

Hint 3 — second domino

Next, the 2–6 goes at row 2, column 2 → row 2, 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: = must all be the same number; >2 must add up to more than 2; 9 must add up to 9.

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

Hard Pips answer — November 14, 2025

5×6 board · 11 dominoes · 11 regions · by Rodolfo Kurchan

Hard today is a 5×6 board with 11 dominoes and 11 regions. You're working with 9 sum regions, 1 "all equal", 1 free region. The single cell at row 1, column 4 is marked 5, so it can only hold a 5. After that, the single cell at row 1, column 5 is marked 5, so it can only hold a 5. Rack check: 3 doubles, 6 blank halves.

Hints

Hint 1 — where to start

The single cell at row 1, column 4 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 5–0 domino with its 5 at row 1, column 4 and its 0 at row 1, column 3.

Hint 3 — second domino

Next, the 5–4 goes at row 1, column 5 → row 1, column 6. 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; 5 must add up to 5.

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

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