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NYT Pips answers — Friday, January 16, 2026

Pips #152 · 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.

Jump to: Easy · Medium · Hard

Easy Pips answer — January 16, 2026

4×5 board · 6 dominoes · 7 regions · by Ian Livengood

The easy board runs 4 rows by 5 columns: 6 dominoes to place across 7 regions. The mix is 2 sum regions, 3 "all equal", 2 free regions. Best foothold — the single cell at row 1, column 3 is marked 3, so it can only hold a 3. After that, the single cell at row 2, column 1 is marked 3, so it can only hold a 3. Rack check: 2 doubles, 3 blank halves, 1 six.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 4–3 goes at row 3, column 1 → row 2, column 1. 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; 3 must add up to 3.

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

Medium Pips answer — January 16, 2026

6×5 board · 7 dominoes · 8 regions · by Rodolfo Kurchan

The medium board runs 6 rows by 5 columns: 7 dominoes to place across 8 regions. The mix is 1 sum region, 1 "less than", 1 "greater than", 4 free regions. Rack check: 3 doubles, 5 sixes.

Hints

Hint 1 — where to start

Start with the smallest regions — they have the fewest ways to be filled.

Hint 2 — first domino

Place the 4–2 domino with its 4 at row 4, column 5 and its 2 at row 4, column 4.

Hint 3 — second domino

Next, the 3–3 goes at row 3, column 3 → row 4, column 3. 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 different; <11 must add up to less than 11; 7 must add up to 7; >4 must add up to more than 4.

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

Hard Pips answer — January 16, 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 January 16, 2026. You're working with 8 sum regions, 1 "all equal". The single cell at row 1, column 6 is marked 1, so it can only hold a 1. After that, the single cell at row 4, column 3 is marked 1, so it can only hold a 1. Rack check: 4 doubles, 3 blank halves, 4 sixes.

Hints

Hint 1 — where to start

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

Hint 2 — first domino

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

Hint 3 — second domino

Next, the 1–4 goes at row 4, column 3 → row 4, 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; 1 must add up to 1; 24 must add up to 24; = must all be the same number; ≠ must all be different.

10
1
24
=
1
≠
20
0
5
5
Solution as text
  1. 4–6 — 4 at row 1, column 3, 6 at row 2, column 3
  2. 3–2 — 3 at row 8, column 4, 2 at row 8, column 3
  3. 5–0 — 5 at row 7, column 2, 0 at row 7, column 1
  4. 3–3 — 3 at row 1, column 4, 3 at row 1, column 5
  5. 4–5 — 4 at row 5, column 4, 5 at row 5, column 3
  6. 1–4 — 1 at row 4, column 3, 4 at row 4, column 4
  7. 5–5 — 5 at row 6, column 3, 5 at row 7, column 3
  8. 4–0 — 4 at row 7, column 6, 0 at row 7, column 7
  9. 3–1 — 3 at row 5, column 7, 1 at row 6, column 7
  10. 6–6 — 6 at row 2, column 4, 6 at row 2, column 5
  11. 2–5 — 2 at row 4, column 7, 5 at row 4, column 8
  12. 6–1 — 6 at row 2, column 6, 1 at row 1, column 6
  13. 4–4 — 4 at row 3, column 3, 4 at row 3, column 4
  14. 3–0 — 3 at row 8, column 2, 0 at row 8, column 1
  15. 4–2 — 4 at row 6, column 4, 2 at row 7, column 4

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