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NYT Pips answers — Tuesday, November 18, 2025

Pips #93 · 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.

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Easy Pips answer — November 18, 2025

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

4 dominoes, 4 regions, a 3×3 grid — that's the easy puzzle for November 18, 2025. You're working with 1 sum region, 2 "all equal", 1 free region. The way in: the "=" region starting at row 1, column 3 has 3 cells that must all match — every 0 in the puzzle should come here.. After that, the "=" pair at row 1, column 1 needs two identical pips — a double, or matching halves from two dominoes.. Rack check: 2 doubles, 3 blank halves, 2 sixes.

Hints

Hint 1 — where to start

The "=" region starting at row 1, column 3 has 3 cells that must all match — every 0 in the puzzle should come here.

Hint 2 — first domino

Place the 0–6 domino with its 0 at row 1, column 3 and its 6 at row 1, column 2.

Hint 3 — second domino

Next, the 2–6 goes at row 2, column 1 → row 1, 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; 8 must add up to 8.

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

Medium Pips answer — November 18, 2025

4×5 board · 8 dominoes · 9 regions · by Ian Livengood

8 dominoes, 9 regions, a 4×5 grid — that's the medium puzzle for November 18, 2025. Rules on the board: 4 sum regions, 3 "all equal", 2 free regions. Best foothold — the single cell at row 1, column 5 is marked 1, so it can only hold a 1. After that, the single cell at row 4, column 4 is marked 4, so it can only hold a 4. Rack check: 1 double, 2 blank halves, 2 sixes.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 4–3 goes at row 4, column 4 → 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: 5 must add up to 5; 1 must add up to 1; = must all be the same number; 11 must add up to 11; 4 must add up to 4.

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

Hard Pips answer — November 18, 2025

7×4 board · 9 dominoes · 8 regions · by Rodolfo Kurchan

The hard board runs 7 rows by 4 columns: 9 dominoes to place across 8 regions. The mix is 6 sum regions, 1 "all equal", 1 free region. The single cell at row 1, column 2 is marked 3, so it can only hold a 3. After that, the "=" pair at row 7, column 1 needs two identical pips — a double, or matching halves from two dominoes.. Rack check: 3 doubles, 2 sixes.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 5–5 goes at row 7, column 1 → row 7, 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: 9 must add up to 9; 3 must add up to 3; = must all be the same number.

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

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