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NYT Pips answers — Thursday, November 20, 2025

Pips #95 · Edited by Ian Livengood · Easy by Ian Livengood · Medium by Ian Livengood · 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 20, 2025

3×4 board · 5 dominoes · 6 regions · by Ian Livengood

The easy board runs 3 rows by 4 columns: 5 dominoes to place across 6 regions. The mix is 2 sum regions, 1 "all equal", 1 "less than", 2 free regions. The single cell at row 1, column 1 is marked 5, so it can only hold a 5. After that, the "=" region starting at row 1, column 2 has 3 cells that must all match — every 0 in the puzzle should come here.. Rack check: 3 doubles, 3 blank halves.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 2–2 goes at row 2, 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: 5 must add up to 5; = must all be the same number; <8 must add up to less than 8; 3 must add up to 3.

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

Medium Pips answer — November 20, 2025

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

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

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 6–5 goes at row 3, column 1 → 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: = must all be the same number; 0 must add up to 0; ≠ must all be different; >5 must add up to more than 5; 5 must add up to 5.

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

Hard Pips answer — November 20, 2025

5×7 board · 12 dominoes · 9 regions · by Rodolfo Kurchan

The hard board runs 5 rows by 7 columns: 12 dominoes to place across 9 regions. Rules on the board: 3 sum regions, 4 "all equal", 2 free regions. The single cell at row 5, column 6 is marked 3, so it can only hold a 3. After that, three cells adding to 18 at row 1, column 2 needs big pips — at least one 6.. Rack check: 3 doubles, 1 blank half, 5 sixes.

Hints

Hint 1 — where to start

The single cell at row 5, column 6 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 3–2 domino with its 3 at row 5, column 6 and its 2 at row 4, column 6.

Hint 3 — second domino

Next, the 6–3 goes at row 1, column 2 → 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: 18 must add up to 18; = must all be the same number; 3 must add up to 3.

18
=
=
=
3
=
3
Solution as text
  1. 0–4 — 0 at row 5, column 2, 4 at row 5, column 3
  2. 5–5 — 5 at row 1, column 6, 5 at row 1, column 7
  3. 1–1 — 1 at row 3, column 2, 1 at row 4, column 2
  4. 6–5 — 6 at row 2, column 5, 5 at row 2, column 6
  5. 4–4 — 4 at row 3, column 3, 4 at row 4, column 3
  6. 1–6 — 1 at row 3, column 1, 6 at row 2, column 1
  7. 3–2 — 3 at row 5, column 6, 2 at row 4, column 6
  8. 6–3 — 6 at row 1, column 2, 3 at row 1, column 3
  9. 2–1 — 2 at row 4, column 7, 1 at row 5, column 7
  10. 6–4 — 6 at row 2, column 2, 4 at row 2, column 3
  11. 5–2 — 5 at row 2, column 7, 2 at row 3, column 7
  12. 6–2 — 6 at row 3, column 5, 2 at row 3, column 6

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