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NYT Pips answers — Friday, August 29, 2025

Pips #12 · Edited by Ian Livengood · Easy by Rodolfo Kurchan · 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.

Jump to: Easy · Medium · Hard

Easy Pips answer — August 29, 2025

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

4 dominoes, 5 regions, a 3×5 grid — that's the easy puzzle for August 29, 2025. Rules on the board: 2 sum regions, 1 "all equal", 2 free regions. Best foothold — the single cell at row 2, column 4 is marked 3, so it can only hold a 3. After that, the single cell at row 3, column 5 is marked 4, so it can only hold a 4. Rack check: 1 double, 1 blank half.

Hints

Hint 1 — where to start

The single cell at row 2, column 4 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 3–1 domino with its 3 at row 2, column 4 and its 1 at row 1, column 4.

Hint 3 — second domino

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

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: 3 must add up to 3; = must all be the same number; 4 must add up to 4.

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

Medium Pips answer — August 29, 2025

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

The medium board runs 3 rows by 6 columns: 6 dominoes to place across 6 regions. The mix is 5 sum regions, 1 free region. Best foothold — the two-cell region at row 1, column 3 adds to 12: both cells must be 6. After that, the two-cell region at row 3, column 3 adds to 0: both cells must be 0. Rack check: 1 double, 3 blank halves, 2 sixes.

Hints

Hint 1 — where to start

The two-cell region at row 1, column 3 adds to 12: both cells must be 6. That's the 6–6 domino, or two 6s from different dominoes.

Hint 2 — first domino

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

Hint 3 — second domino

Next, the 0–4 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: 12 must add up to 12; 4 must add up to 4; 8 must add up to 8; 0 must add up to 0.

12
4
8
4
0
Solution as text
  1. 2–0 — 2 at row 2, column 4, 0 at row 3, column 4
  2. 0–6 — 0 at row 1, column 5, 6 at row 1, column 4
  3. 0–4 — 0 at row 3, column 3, 4 at row 3, column 2
  4. 2–6 — 2 at row 2, column 3, 6 at row 1, column 3
  5. 4–5 — 4 at row 2, column 2, 5 at row 2, column 1
  6. 2–2 — 2 at row 2, column 5, 2 at row 2, column 6

Hard Pips answer — August 29, 2025

10×2 board · 10 dominoes · 10 regions · by Rodolfo Kurchan

The hard board runs 10 rows by 2 columns: 10 dominoes to place across 10 regions. The mix is 6 sum regions, 2 "all equal", 2 free regions. Best foothold — the single cell at row 9, column 1 is marked 3, so it can only hold a 3. After that, the two-cell region at row 4, column 2 adds to 12: both cells must be 6. Rack check: 2 doubles, 2 blank halves, 5 sixes.

Hints

Hint 1 — where to start

The single cell at row 9, column 1 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 5–3 domino with its 5 at row 8, column 1 and its 3 at row 9, column 1.

Hint 3 — second domino

Next, the 2–6 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: 9 must add up to 9; = must all be the same number; 12 must add up to 12; 2 must add up to 2; 10 must add up to 10.

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

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