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NYT Pips answers — Thursday, February 26, 2026

Pips #193 · 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 — February 26, 2026

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

5 dominoes, 5 regions, a 3×4 grid — that's the easy puzzle for February 26, 2026. You're working with 2 sum regions, 3 "all equal". The way in: the single cell at row 2, column 1 is marked 3, so it can only hold a 3. After that, the "=" region starting at row 3, column 1 has 3 cells that must all match — every 0 in the puzzle should come here.. Rack check: 2 doubles, 3 blank halves.

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

Hint 3 — second domino

Next, the 0–0 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: = must all be the same number; 7 must add up to 7; 3 must add up to 3.

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

Medium Pips answer — February 26, 2026

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

Medium today is a 3×6 board with 7 dominoes and 9 regions. The mix is 3 sum regions, 1 "less than", 1 "greater than", 3 free regions. The two-cell region at row 1, column 2 adds to 11: the only split is 5 + 6.. After that, the "<2" region at row 3, column 6 leaves almost no room: only 0s and a single 1 can go here.. Rack check: 4 doubles, 4 blank halves, 1 six.

Hints

Hint 1 — where to start

The two-cell region at row 1, column 2 adds to 11: the only split is 5 + 6.

Hint 2 — first domino

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

Hint 3 — second domino

Next, the 0–0 goes at row 2, column 6 → row 3, 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: 10 must add up to 10; 11 must add up to 11; ≠ must all be different; 8 must add up to 8; >2 must add up to more than 2.

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

Hard Pips answer — February 26, 2026

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

Hard today is a 10×8 board with 16 dominoes and 16 regions. Rules on the board: 9 sum regions, 3 "all equal", 1 "less than", 1 "greater than", 2 free regions. The way in: the single cell at row 8, column 5 is marked 4, so it can only hold a 4. After that, the two-cell region at row 3, column 6 adds to 12: both cells must be 6. Rack check: 6 doubles, 4 blank halves, 5 sixes.

Hints

Hint 1 — where to start

The single cell at row 8, column 5 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 0–4 domino with its 0 at row 8, column 6 and its 4 at row 8, column 5.

Hint 3 — second domino

Next, the 2–6 goes at row 2, column 6 → row 3, 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: 10 must add up to 10; <2 must add up to less than 2; 6 must add up to 6; = must all be the same number; 2 must add up to 2.

10
<2
6
=
2
=
12
9
=
17
4
1
>4
4
Solution as text
  1. 2–6 — 2 at row 2, column 6, 6 at row 3, column 6
  2. 5–5 — 5 at row 3, column 1, 5 at row 4, column 1
  3. 2–1 — 2 at row 2, column 7, 1 at row 2, column 8
  4. 6–6 — 6 at row 6, column 5, 6 at row 6, column 6
  5. 0–4 — 0 at row 8, column 6, 4 at row 8, column 5
  6. 3–3 — 3 at row 2, column 4, 3 at row 3, column 4
  7. 0–6 — 0 at row 1, column 3, 6 at row 1, column 2
  8. 2–4 — 2 at row 2, column 1, 4 at row 1, column 1
  9. 0–0 — 0 at row 9, column 6, 0 at row 10, column 6
  10. 4–1 — 4 at row 4, column 8, 1 at row 3, column 8
  11. 3–5 — 3 at row 10, column 3, 5 at row 9, column 3
  12. 1–1 — 1 at row 10, column 4, 1 at row 10, column 5
  13. 6–5 — 6 at row 4, column 6, 5 at row 4, column 7
  14. 5–4 — 5 at row 6, column 4, 4 at row 6, column 3
  15. 5–1 — 5 at row 4, column 2, 1 at row 4, column 3
  16. 4–4 — 4 at row 7, column 3, 4 at row 8, column 3

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