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NYT Pips answers — Thursday, May 21, 2026

Pips #277 · 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 — May 21, 2026

3×6 board · 5 dominoes · 6 regions · by Ian Livengood

The easy board runs 3 rows by 6 columns: 5 dominoes to place across 6 regions. The mix is 4 sum regions, 2 "greater than". The single cell marked ">5" at row 1, column 1 must be a 6.. After that, the single cell marked ">5" at row 1, column 6 must be a 6.. Rack check: 3 doubles, 3 sixes.

Hints

Hint 1 — where to start

The single cell marked ">5" at row 1, column 1 must be a 6.

Hint 2 — first domino

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

Hint 3 — second domino

Next, the 6–6 goes at row 1, column 6 → row 2, 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: >5 must add up to more than 5; 8 must add up to 8; 5 must add up to 5; 6 must add up to 6.

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

Medium Pips answer — May 21, 2026

7×4 board · 7 dominoes · 8 regions · by Ian Livengood

Medium today is a 7×4 board with 7 dominoes and 8 regions. Rules on the board: 3 sum regions, 3 "all equal", 2 "less than". Best foothold — the single cell at row 7, column 1 is marked 3, so it can only hold a 3. After that, the "<2" region at row 1, column 3 leaves almost no room: only 0s and a single 1 can go here.. Rack check: 2 blank halves, 2 sixes.

Hints

Hint 1 — where to start

The single cell at row 7, 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 3–5 domino with its 3 at row 7, column 1 and its 5 at row 7, column 2.

Hint 3 — second domino

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

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

Hard Pips answer — May 21, 2026

5×6 board · 15 dominoes · 29 regions · by Rodolfo Kurchan

The hard board runs 5 rows by 6 columns: 15 dominoes to place across 29 regions. The mix is 26 sum regions, 1 "all equal", 2 free regions. The single cell at row 1, column 1 is marked 0, so it can only hold a 0. After that, the single cell at row 1, column 2 is marked 2, so it can only hold a 2. Rack check: 5 blank halves.

Hints

Hint 1 — where to start

The single cell at row 1, column 1 is marked 0, so it can only hold a 0. Find the domino with a 0 that fits there.

Hint 2 — first domino

Place the 0–4 domino with its 0 at row 1, column 1 and its 4 at row 2, column 1.

Hint 3 — second domino

Next, the 2–4 goes at row 1, column 2 → row 2, 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: 0 must add up to 0; 2 must add up to 2; 1 must add up to 1; 5 must add up to 5; 4 must add up to 4.

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

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