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NYT Pips answers — Monday, May 11, 2026

Pips #267 · 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.

Jump to: Easy · Medium · Hard

Easy Pips answer — May 11, 2026

3×3 board · 4 dominoes · 5 regions · by Ian Livengood

The easy board runs 3 rows by 3 columns: 4 dominoes to place across 5 regions. You're working with 2 sum regions, 1 "less than", 2 free regions. The way in: the pair at row 2, column 1 adds to 4; list the ways to make 4 with the rack you have.. After that, the pair at row 3, column 2 adds to 10; list the ways to make 10 with the rack you have.. Rack check: 1 double, 1 blank half, 2 sixes.

Hints

Hint 1 — where to start

The pair at row 2, column 1 adds to 4; list the ways to make 4 with the rack you have.

Hint 2 — first domino

Place the 3–4 domino with its 3 at row 1, column 1 and its 4 at row 2, column 1.

Hint 3 — second domino

Next, the 0–4 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: <10 must add up to less than 10; 4 must add up to 4; 10 must add up to 10.

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

Medium Pips answer — May 11, 2026

4×5 board · 7 dominoes · 9 regions · by Ian Livengood

7 dominoes, 9 regions, a 4×5 grid — that's the medium puzzle for May 11, 2026. You're working with 4 sum regions, 1 "all equal", 2 "less than", 2 free regions. The way in: the single cell at row 1, column 3 is marked 4, so it can only hold a 4. After that, the "<1" region at row 2, column 4 leaves almost no room: every cell must be 0.. Rack check: 3 blank halves, 1 six.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

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

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

Hard Pips answer — May 11, 2026

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

15 dominoes, 10 regions, a 8×8 grid — that's the hard puzzle for May 11, 2026. You're working with 4 sum regions, 4 "all equal", 1 free region. The way in: the single cell at row 1, column 6 is marked 2, so it can only hold a 2. After that, the single cell at row 8, column 2 is marked 2, so it can only hold a 2. Rack check: 3 doubles, 1 blank half, 4 sixes.

Hints

Hint 1 — where to start

The single cell at row 1, column 6 is marked 2, so it can only hold a 2. Find the domino with a 2 that fits there.

Hint 2 — first domino

Place the 1–2 domino with its 1 at row 1, column 5 and its 2 at row 1, column 6.

Hint 3 — second domino

Next, the 6–2 goes at row 8, column 1 → row 8, 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; 2 must add up to 2; ≠ must all be different; 3 must add up to 3.

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

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