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NYT Pips answers — Sunday, August 31, 2025

Pips #14 · Edited by Ian Livengood · Easy by Rodolfo Kurchan · Medium by Rodolfo Kurchan · Hard by Ian Livengood

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 — August 31, 2025

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

Easy today is a 3×4 board with 4 dominoes and 5 regions. The mix is 1 sum region, 2 "all equal", 2 free regions. The "=" pair at row 1, column 3 needs two identical pips — a double, or matching halves from two dominoes.. After that, the "=" pair at row 2, column 4 needs two identical pips — a double, or matching halves from two dominoes.. Rack check: 1 blank half, 1 six.

Hints

Hint 1 — where to start

The "=" pair at row 1, column 3 needs two identical pips — a double, or matching halves from two dominoes.

Hint 2 — first domino

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

Hint 3 — second domino

Next, the 2–1 goes at row 2, column 4 → row 1, column 4. 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; 10 must add up to 10.

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

Medium Pips answer — August 31, 2025

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

The medium board runs 3 rows by 4 columns: 4 dominoes to place across 5 regions. Rules on the board: 1 "all equal", 3 "greater than", 1 free region. The way in: the "=" pair at row 3, column 1 needs two identical pips — a double, or matching halves from two dominoes.. Rack check: 1 blank half, 1 six.

Hints

Hint 1 — where to start

The "=" pair at row 3, column 1 needs two identical pips — a double, or matching halves from two dominoes.

Hint 2 — first domino

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

Hint 3 — second domino

Next, the 5–4 goes at row 2, column 3 → row 2, column 4. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: >6 must add up to more than 6; >4 must add up to more than 4; = must all be the same number.

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

Hard Pips answer — August 31, 2025

7×7 board · 12 dominoes · 10 regions · by Ian Livengood

12 dominoes, 10 regions, a 7×7 grid — that's the hard puzzle for August 31, 2025. You're working with 6 sum regions, 2 "all equal", 2 free regions. The way in: the single cell at row 1, column 7 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: 5 doubles, 2 blank halves, 4 sixes.

Hints

Hint 1 — where to start

The single cell at row 1, column 7 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 1–0 domino with its 1 at row 1, column 6 and its 0 at row 1, column 7.

Hint 3 — second domino

Next, the 5–6 goes at row 4, column 1 → row 4, 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: 13 must add up to 13; = must all be the same number; 0 must add up to 0; 24 must add up to 24; 4 must add up to 4.

13
=
0
24
4
5
=
0
Solution as text
  1. 2–4 — 2 at row 5, column 2, 4 at row 5, column 3
  2. 3–3 — 3 at row 1, column 4, 3 at row 1, column 5
  3. 2–2 — 2 at row 4, column 3, 2 at row 4, column 4
  4. 4–0 — 4 at row 6, column 3, 0 at row 7, column 3
  5. 4–6 — 4 at row 2, column 4, 6 at row 2, column 3
  6. 2–1 — 2 at row 3, column 4, 1 at row 3, column 5
  7. 3–4 — 3 at row 4, column 5, 4 at row 5, column 5
  8. 5–6 — 5 at row 4, column 1, 6 at row 4, column 2
  9. 1–1 — 1 at row 2, column 5, 1 at row 2, column 6
  10. 4–4 — 4 at row 5, column 4, 4 at row 6, column 4
  11. 6–6 — 6 at row 3, column 2, 6 at row 3, column 3
  12. 1–0 — 1 at row 1, column 6, 0 at row 1, column 7

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