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NYT Pips answers — Friday, September 5, 2025

Pips #19 · Edited by Ian Livengood · Easy by Heidi Erwin · Medium by Heidi Erwin · 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 — September 5, 2025

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

The easy board runs 4 rows by 3 columns: 4 dominoes to place across 5 regions. The mix is 3 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 single cell at row 4, column 2 is marked 3, so it can only hold a 3. Rack check: 1 double.

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 2 and its 3 at row 2, column 2.

Hint 3 — second domino

Next, the 2–3 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: 3 must add up to 3; = must all be the same number; 7 must add up to 7.

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

Medium Pips answer — September 5, 2025

5×4 board · 7 dominoes · 7 regions · by Heidi Erwin

7 dominoes, 7 regions, a 5×4 grid — that's the medium puzzle for September 5, 2025. You're working with 3 sum regions, 2 "all equal", 1 "less than", 1 free region. The single cell at row 1, column 2 is marked 5, so it can only hold a 5. After that, the single cell at row 4, column 4 is marked 5, so it can only hold a 5. Rack check: 3 doubles, 1 six.

Hints

Hint 1 — where to start

The single cell at row 1, column 2 is marked 5, so it can only hold a 5. Find the domino with a 5 that fits there.

Hint 2 — first domino

Place the 2–5 domino with its 2 at row 1, column 1 and its 5 at row 1, column 2.

Hint 3 — second domino

Next, the 4–5 goes at row 4, column 3 → row 4, 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: <4 must add up to less than 4; 5 must add up to 5; = must all be the same number; 6 must add up to 6.

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

Hard Pips answer — September 5, 2025

10×8 board · 15 dominoes · 12 regions · by Rodolfo Kurchan

15 dominoes, 12 regions, a 10×8 grid — that's the hard puzzle for September 5, 2025. Rules on the board: 5 sum regions, 5 "all equal", 1 free region. Best foothold — the single cell at row 4, column 4 is marked 6, so it can only hold a 6. After that, the "=" region starting at row 5, column 3 has 3 cells that must all match — every 4 in the puzzle should come here.. Rack check: 4 doubles, 5 blank halves, 4 sixes.

Hints

Hint 1 — where to start

The single cell at row 4, column 4 is marked 6, so it can only hold a 6. Find the domino with a 6 that fits there.

Hint 2 — first domino

Place the 2–6 domino with its 2 at row 4, column 3 and its 6 at row 4, column 4.

Hint 3 — second domino

Next, the 4–1 goes at row 5, column 3 → row 5, 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 different; = must all be the same number; 2 must add up to 2; 6 must add up to 6; 0 must add up to 0.

≠
=
2
=
=
6
2
=
=
0
18
Solution as text
  1. 4–0 — 4 at row 7, column 3, 0 at row 7, column 4
  2. 1–5 — 1 at row 5, column 1, 5 at row 6, column 1
  3. 5–5 — 5 at row 7, column 2, 5 at row 8, column 2
  4. 3–3 — 3 at row 3, column 2, 3 at row 4, column 2
  5. 2–6 — 2 at row 4, column 3, 6 at row 4, column 4
  6. 4–5 — 4 at row 6, column 3, 5 at row 6, column 2
  7. 6–6 — 6 at row 10, column 7, 6 at row 10, column 8
  8. 3–1 — 3 at row 2, column 5, 1 at row 2, column 4
  9. 5–0 — 5 at row 8, column 3, 0 at row 8, column 4
  10. 2–3 — 2 at row 1, column 7, 3 at row 1, column 8
  11. 2–5 — 2 at row 1, column 6, 5 at row 1, column 5
  12. 0–0 — 0 at row 9, column 4, 0 at row 9, column 5
  13. 2–1 — 2 at row 3, column 3, 1 at row 3, column 4
  14. 4–1 — 4 at row 5, column 3, 1 at row 5, column 2
  15. 6–0 — 6 at row 10, column 6, 0 at row 10, column 5

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