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NYT Pips answers — Friday, July 3, 2026

Pips #320 · Edited by Ian Livengood · Easy by Ian Livengood · Medium by Ian Livengood · 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.

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Easy Pips answer — July 3, 2026

4×4 board · 5 dominoes · 7 regions · by Ian Livengood

Easy today is a 4×4 board with 5 dominoes and 7 regions. Rules on the board: 3 sum regions, 2 "less than", 1 "greater than", 1 free region. Best foothold — the single cell marked ">5" at row 4, column 4 must be a 6.. After that, the "<2" region at row 1, column 2 leaves almost no room: only 0s and a single 1 can go here.. Rack check: 1 blank half, 2 sixes.

Hints

Hint 1 — where to start

The single cell marked ">5" at row 4, column 4 must be a 6.

Hint 2 — first domino

Place the 5–6 domino with its 5 at row 4, column 3 and its 6 at row 4, column 4.

Hint 3 — second domino

Next, the 1–6 goes at row 1, column 2 → row 1, 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: 10 must add up to 10; <2 must add up to less than 2; 3 must add up to 3; >5 must add up to more than 5.

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

Medium Pips answer — July 3, 2026

3×9 board · 9 dominoes · 10 regions · by Ian Livengood

The medium board runs 3 rows by 9 columns: 9 dominoes to place across 10 regions. The mix is 3 sum regions, 5 "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 2, column 7 is marked 4, so it can only hold a 4. Rack check: 3 doubles, 1 blank half, 4 sixes.

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

Hint 3 — second domino

Next, the 4–3 goes at row 2, column 7 → row 3, column 7. 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; 5 must add up to 5; <2 must add up to less than 2; 4 must add up to 4; 8 must add up to 8.

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

Hard Pips answer — July 3, 2026

10×8 board · 16 dominoes · 19 regions · by Rodolfo Kurchan

16 dominoes, 19 regions, a 10×8 grid — that's the hard puzzle for July 3, 2026. You're working with 12 sum regions, 3 "all equal", 4 free regions. The way in: 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 6, column 6 is marked 1, so it can only hold a 1. Rack check: 3 doubles, 4 blank halves, 4 sixes.

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 5–3 domino with its 5 at row 3, column 4 and its 3 at row 2, column 4.

Hint 3 — second domino

Next, the 3–1 goes at row 6, column 7 → row 6, column 6. 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; 7 must add up to 7; = must all be the same number; 10 must add up to 10; 12 must add up to 12.

3
7
=
10
=
12
2
6
5
1
3
5
=
12
10
Solution as text
  1. 1–5 — 1 at row 5, column 8, 5 at row 6, column 8
  2. 4–4 — 4 at row 9, column 4, 4 at row 10, column 4
  3. 3–1 — 3 at row 6, column 7, 1 at row 6, column 6
  4. 0–2 — 0 at row 3, column 7, 2 at row 3, column 6
  5. 5–3 — 5 at row 3, column 4, 3 at row 2, column 4
  6. 2–2 — 2 at row 1, column 7, 2 at row 1, column 8
  7. 6–4 — 6 at row 8, column 3, 4 at row 8, column 4
  8. 5–5 — 5 at row 9, column 2, 5 at row 10, column 2
  9. 3–2 — 3 at row 6, column 2, 2 at row 6, column 3
  10. 0–1 — 0 at row 3, column 8, 1 at row 4, column 8
  11. 6–3 — 6 at row 8, column 2, 3 at row 7, column 2
  12. 0–4 — 0 at row 3, column 1, 4 at row 2, column 1
  13. 5–6 — 5 at row 4, column 4, 6 at row 4, column 3
  14. 3–4 — 3 at row 6, column 4, 4 at row 7, column 4
  15. 0–6 — 0 at row 4, column 1, 6 at row 4, column 2
  16. 5–4 — 5 at row 2, column 6, 4 at row 1, column 6

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