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NYT Pips answers — Wednesday, May 27, 2026

Pips #283 · 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.

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Easy Pips answer — May 27, 2026

4×4 board · 5 dominoes · 6 regions · by Ian Livengood

5 dominoes, 6 regions, a 4×4 grid — that's the easy puzzle for May 27, 2026. You're working with 2 sum regions, 1 "all equal", 1 "less than", 1 "greater than", 1 free region. The single cell at row 2, column 1 is marked 3, so it can only hold a 3. After that, the single cell at row 2, column 4 is marked 3, so it can only hold a 3. Rack check: 2 blank halves, 1 six.

Hints

Hint 1 — where to start

The single cell at row 2, column 1 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 3–0 domino with its 3 at row 2, column 1 and its 0 at row 1, column 1.

Hint 3 — second domino

Next, the 3–5 goes at row 2, column 4 → row 1, 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: <3 must add up to less than 3; >15 must add up to more than 15; 3 must add up to 3; = must all be the same number.

<3
>15
3
3
=
Solution as text
  1. 3–5 — 3 at row 2, column 4, 5 at row 1, column 4
  2. 3–0 — 3 at row 2, column 1, 0 at row 1, column 1
  3. 5–2 — 5 at row 1, column 3, 2 at row 1, column 2
  4. 4–0 — 4 at row 4, column 3, 0 at row 4, column 4
  5. 4–6 — 4 at row 3, column 3, 6 at row 2, column 3

Medium Pips answer — May 27, 2026

3×6 board · 7 dominoes · 9 regions · by Rodolfo Kurchan

7 dominoes, 9 regions, a 3×6 grid — that's the medium puzzle for May 27, 2026. You're working with 5 sum regions, 1 "greater than", 3 free regions. The way in: the pair at row 1, column 2 adds to 10; list the ways to make 10 with the rack you have.. After that, the pair at row 1, column 3 adds to 7; list the ways to make 7 with the rack you have.. Rack check: 2 blank halves, 2 sixes.

Hints

Hint 1 — where to start

The pair at row 1, column 2 adds to 10; list the ways to make 10 with the rack you have.

Hint 2 — first domino

Place the 6–4 domino with its 6 at row 1, column 3 and its 4 at row 1, column 2.

Hint 3 — second domino

Next, the 5–4 goes at row 1, column 5 → row 1, 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: 10 must add up to 10; 7 must add up to 7; 8 must add up to 8; >0 must add up to more than 0; 6 must add up to 6.

10
7
8
8
>0
6
Solution as text
  1. 1–4 — 1 at row 2, column 3, 4 at row 2, column 4
  2. 0–3 — 0 at row 2, column 6, 3 at row 2, column 5
  3. 6–4 — 6 at row 1, column 3, 4 at row 1, column 2
  4. 1–2 — 1 at row 3, column 2, 2 at row 3, column 3
  5. 5–4 — 5 at row 1, column 5, 4 at row 1, column 4
  6. 6–2 — 6 at row 2, column 2, 2 at row 2, column 1
  7. 4–0 — 4 at row 3, column 4, 0 at row 3, column 5

Hard Pips answer — May 27, 2026

5×6 board · 15 dominoes · 30 regions · by Rodolfo Kurchan

The hard board runs 5 rows by 6 columns: 15 dominoes to place across 30 regions. You're working with 27 sum regions, 3 free regions. 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 2 is marked 4, so it can only hold a 4. Rack check: 2 doubles, 3 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 4–4 domino with its 4 at row 1, column 1 and its 4 at row 1, column 2.

Hint 3 — second domino

Next, the 3–4 goes at row 1, column 3 → row 1, 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; 3 must add up to 3; 1 must add up to 1; 5 must add up to 5; 0 must add up to 0.

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

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