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NYT Pips answers — Monday, February 16, 2026

Pips #183 · 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 16, 2026

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

Easy today is a 4×4 board with 5 dominoes and 6 regions. Rules on the board: 2 sum regions, 1 "all equal", 3 free regions. Best foothold — the "=" region starting at row 1, column 3 has 3 cells that must all match — every 1 in the puzzle should come here.. After that, the pair at row 3, column 2 adds to 9; list the ways to make 9 with the rack you have.. Rack check: 3 doubles, 2 blank halves, 2 sixes.

Hints

Hint 1 — where to start

The "=" region starting at row 1, column 3 has 3 cells that must all match — every 1 in the puzzle should come here.

Hint 2 — first domino

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

Hint 3 — second domino

Next, the 6–5 goes at row 3, column 3 → 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; 9 must add up to 9; 6 must add up to 6.

=
9
6
Solution as text
  1. 6–1 — 6 at row 2, column 1, 1 at row 2, column 2
  2. 0–0 — 0 at row 3, column 4, 0 at row 4, column 4
  3. 4–4 — 4 at row 4, column 1, 4 at row 4, column 2
  4. 6–5 — 6 at row 3, column 3, 5 at row 3, column 2
  5. 1–1 — 1 at row 1, column 3, 1 at row 2, column 3

Medium Pips answer — February 16, 2026

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

7 dominoes, 7 regions, a 5×5 grid — that's the medium puzzle for February 16, 2026. The mix is 2 sum regions, 2 "less than", 1 "greater than", 1 free region. Best foothold — the single cell marked ">5" at row 5, column 5 must be a 6.. After that, the pair at row 1, column 1 adds to 8; list the ways to make 8 with the rack you have.. Rack check: 3 doubles, 3 blank halves, 3 sixes.

Hints

Hint 1 — where to start

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

Hint 2 — first domino

Place the 6–4 domino with its 6 at row 5, column 5 and its 4 at row 5, column 4.

Hint 3 — second domino

Next, the 4–4 goes at row 1, column 1 → 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: 8 must add up to 8; <3 must add up to less than 3; <5 must add up to less than 5; ≠ must all be different; >5 must add up to more than 5.

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

Hard Pips answer — February 16, 2026

9×9 board · 12 dominoes · 9 regions · by Rodolfo Kurchan

Hard today is a 9×9 board with 12 dominoes and 9 regions. Rules on the board: 4 sum regions, 3 "all equal", 1 free region. Best foothold — the two-cell region at row 4, column 4 adds to 1: the only split is 0 + 1.. After that, the two-cell region at row 6, column 4 adds to 11: the only split is 5 + 6.. Rack check: 1 double, 3 blank halves, 2 sixes.

Hints

Hint 1 — where to start

The two-cell region at row 4, column 4 adds to 1: the only split is 0 + 1.

Hint 2 — first domino

Place the 5–0 domino with its 5 at row 4, column 5 and its 0 at row 4, column 4.

Hint 3 — second domino

Next, the 1–6 goes at row 5, column 4 → row 6, 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: = must all be the same number; 2 must add up to 2; 1 must add up to 1; 7 must add up to 7; 11 must add up to 11.

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

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