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NYT Pips answers — Wednesday, September 17, 2025

Pips #31 · Edited by Ian Livengood · Easy by Heidi Erwin · Medium by Rodolfo Kurchan · 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 17, 2025

3×4 board · 4 dominoes · 5 regions · by Heidi Erwin

4 dominoes, 5 regions, a 3×4 grid — that's the easy puzzle for September 17, 2025. You're working with 3 sum regions, 1 "all equal", 1 free region. The single cell at row 2, column 4 is marked 1, so it can only hold a 1. After that, the single cell at row 3, column 2 is marked 0, so it can only hold a 0. Rack check: 1 blank half.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 3–0 goes at row 2, column 2 → 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: = must all be the same number; 9 must add up to 9; 1 must add up to 1; 0 must add up to 0.

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

Medium Pips answer — September 17, 2025

5×5 board · 7 dominoes · 10 regions · by Rodolfo Kurchan

7 dominoes, 10 regions, a 5×5 grid — that's the medium puzzle for September 17, 2025. You're working with 6 sum regions, 1 "all equal", 3 free regions. The single cell at row 1, column 1 is marked 0, so it can only hold a 0. After that, the single cell at row 1, column 3 is marked 6, so it can only hold a 6. Rack check: 1 double, 2 blank halves, 1 six.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 1–6 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: 0 must add up to 0; 6 must add up to 6; 3 must add up to 3; = must all be the same number; 8 must add up to 8.

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

Hard Pips answer — September 17, 2025

8×4 board · 9 dominoes · 10 regions · by Heidi Erwin

The hard board runs 8 rows by 4 columns: 9 dominoes to place across 10 regions. The mix is 5 sum regions, 2 free regions. The single cell at row 1, column 1 is marked 0, so it can only hold a 0. After that, the single cell at row 3, column 1 is marked 0, so it can only hold a 0. Rack check: 3 doubles, 6 blank halves.

Hints

Hint 1 — where to start

The single cell at row 1, 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 1, column 1 and its 2 at row 2, column 1.

Hint 3 — second domino

Next, the 0–4 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: 0 must add up to 0; ≠ must all be different.

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

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