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NYT Pips answers — Monday, June 8, 2026

Pips #295 · Edited by Ian Livengood · Easy by Ian Livengood · 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 — June 8, 2026

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

Easy today is a 4×4 board with 5 dominoes and 7 regions. You're working with 3 sum regions, 1 "less than", 3 free regions. The "<1" region at row 4, column 3 leaves almost no room: every cell must be 0.. After that, the pair at row 2, column 1 adds to 8; list the ways to make 8 with the rack you have.. Rack check: 1 blank half, 2 sixes.

Hints

Hint 1 — where to start

The "<1" region at row 4, column 3 leaves almost no room: every cell must be 0.

Hint 2 — first domino

Place the 2–0 domino with its 2 at row 4, column 4 and its 0 at row 4, column 3.

Hint 3 — second domino

Next, the 3–2 goes at row 1, column 1 → row 2, 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: 8 must add up to 8; 10 must add up to 10; 5 must add up to 5; <1 must add up to less than 1.

8
10
5
<1
Solution as text
  1. 6–1 — 6 at row 3, column 1, 1 at row 3, column 2
  2. 5–3 — 5 at row 2, column 4, 3 at row 3, column 4
  3. 2–0 — 2 at row 4, column 4, 0 at row 4, column 3
  4. 3–2 — 3 at row 1, column 1, 2 at row 2, column 1
  5. 6–5 — 6 at row 1, column 3, 5 at row 2, column 3

Medium Pips answer — June 8, 2026

4×5 board · 7 dominoes · 8 regions · by Rodolfo Kurchan

7 dominoes, 8 regions, a 4×5 grid — that's the medium puzzle for June 8, 2026. You're working with 3 sum regions, 2 "all equal", 2 "greater than", 1 free region. The way in: the single cell at row 4, column 2 is marked 5, so it can only hold a 5. After that, the "=" pair at row 3, column 1 needs two identical pips — a double, or matching halves from two dominoes.. Rack check: 1 double, 4 blank halves, 1 six.

Hints

Hint 1 — where to start

The single cell at row 4, 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 0–5 domino with its 0 at row 4, column 1 and its 5 at row 4, column 2.

Hint 3 — second domino

Next, the 0–6 goes at row 3, column 1 → 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: 11 must add up to 11; 7 must add up to 7; = must all be the same number; 5 must add up to 5; >8 must add up to more than 8.

11
7
=
=
5
>8
>0
Solution as text
  1. 0–6 — 0 at row 3, column 1, 6 at row 3, column 2
  2. 0–5 — 0 at row 4, column 1, 5 at row 4, column 2
  3. 4–4 — 4 at row 1, column 4, 4 at row 1, column 5
  4. 3–0 — 3 at row 2, column 4, 0 at row 3, column 4
  5. 1–5 — 1 at row 4, column 5, 5 at row 4, column 4
  6. 0–4 — 0 at row 3, column 3, 4 at row 4, column 3
  7. 2–1 — 2 at row 2, column 3, 1 at row 2, column 2

Hard Pips answer — June 8, 2026

5×7 board · 15 dominoes · 17 regions · by Rodolfo Kurchan

Hard today is a 5×7 board with 15 dominoes and 17 regions. Rules on the board: 10 sum regions, 2 "all equal", 1 "less than", 2 "greater than", 1 free region. The way in: the single cell at row 1, column 1 is marked 4, so it can only hold a 4. After that, the single cell at row 1, column 6 is marked 0, so it can only hold a 0. Rack check: 5 doubles, 6 blank halves, 2 sixes.

Hints

Hint 1 — where to start

The single cell at row 1, column 1 is marked 4, so it can only hold a 4. Find the domino with a 4 that fits there.

Hint 2 — first domino

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

Hint 3 — second domino

Next, the 1–0 goes at row 1, column 5 → row 1, 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: 4 must add up to 4; = must all be the same number; 0 must add up to 0; >4 must add up to more than 4; <4 must add up to less than 4.

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

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