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NYT Pips answers — Thursday, September 4, 2025

Pips #18 · Edited by Ian Livengood · Easy by Heidi Erwin · Medium by Heidi Erwin · Hard by Heidi Erwin

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 — September 4, 2025

4×2 board · 4 dominoes · 4 regions · by Heidi Erwin

Easy today is a 4×2 board with 4 dominoes and 4 regions. Rules on the board: 4 sum regions. The way in: the single cell at row 1, column 2 is marked 0, so it can only hold a 0. After that, the pair at row 3, column 1 adds to 10; list the ways to make 10 with the rack you have.. Rack check: 1 blank half, 1 six.

Hints

Hint 1 — where to start

The single cell at row 1, column 2 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 3–0 domino with its 3 at row 1, column 1 and its 0 at row 1, column 2.

Hint 3 — second domino

Next, the 5–1 goes at row 3, column 1 → 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: 5 must add up to 5; 0 must add up to 0; 10 must add up to 10; 9 must add up to 9.

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

Medium Pips answer — September 4, 2025

4×3 board · 6 dominoes · 6 regions · by Heidi Erwin

6 dominoes, 6 regions, a 4×3 grid — that's the medium puzzle for September 4, 2025. Rules on the board: 1 sum region, 3 "all equal", 2 free regions. Best foothold — the single cell at row 4, column 2 is marked 1, so it can only hold a 1. After that, the "=" region starting at row 1, column 2 has 4 cells that must all match — every 5 in the puzzle should come here.. Rack check: 1 double, 1 blank half.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 2–5 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; 1 must add up to 1.

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

Hard Pips answer — September 4, 2025

5×4 board · 10 dominoes · 10 regions · by Heidi Erwin

The hard board runs 5 rows by 4 columns: 10 dominoes to place across 10 regions. The mix is 4 sum regions, 2 "all equal", 1 "greater than", 3 free regions. The single cell at row 1, column 3 is marked 0, so it can only hold a 0. After that, the two-cell region at row 5, column 2 adds to 0: both cells must be 0. Rack check: 4 blank halves.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 0–1 goes at row 5, column 2 → row 5, 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: 24 must add up to 24; >1 must add up to more than 1; 0 must add up to 0; 4 must add up to 4; = must all be the same number.

24
>1
0
4
=
=
0
Solution as text
  1. 1–4 — 1 at row 3, column 2, 4 at row 2, column 2
  2. 3–4 — 3 at row 4, column 1, 4 at row 4, column 2
  3. 5–3 — 5 at row 2, column 4, 3 at row 3, column 4
  4. 0–1 — 0 at row 5, column 2, 1 at row 5, column 1
  5. 5–1 — 5 at row 2, column 3, 1 at row 3, column 3
  6. 0–3 — 0 at row 1, column 3, 3 at row 1, column 4
  7. 0–5 — 0 at row 3, column 1, 5 at row 2, column 1
  8. 2–5 — 2 at row 1, column 2, 5 at row 1, column 1
  9. 4–0 — 4 at row 4, column 3, 0 at row 5, column 3
  10. 2–4 — 2 at row 5, column 4, 4 at row 4, column 4

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