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NYT Pips answers — Thursday, January 29, 2026

Pips #165 · 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 29, 2026

3×4 board · 5 dominoes · 6 regions · by Ian Livengood

The easy board runs 3 rows by 4 columns: 5 dominoes to place across 6 regions. The mix is 2 sum regions, 2 "greater than", 2 free regions. The single cell marked ">5" at row 1, column 4 must be a 6.. After that, three cells adding to 2 at row 2, column 1 means mostly zeros. Rack check: 2 doubles, 2 blank halves, 1 six.

Hints

Hint 1 — where to start

The single cell marked ">5" at row 1, column 4 must be a 6.

Hint 2 — first domino

Place the 0–6 domino with its 0 at row 1, column 3 and its 6 at row 1, column 4.

Hint 3 — second domino

Next, the 1–1 goes at row 2, column 1 → row 3, 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: >3 must add up to more than 3; 8 must add up to 8; >5 must add up to more than 5; 2 must add up to 2.

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

Medium Pips answer — January 29, 2026

4×5 board · 7 dominoes · 7 regions · by Rodolfo Kurchan

7 dominoes, 7 regions, a 4×5 grid — that's the medium puzzle for January 29, 2026. You're working with 5 sum regions, 1 "all equal", 1 free region. The way in: the "=" pair at row 3, column 4 needs two identical pips — a double, or matching halves from two dominoes.. 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: 2 blank halves, 3 sixes.

Hints

Hint 1 — where to start

The "=" pair at row 3, column 4 needs two identical pips — a double, or matching halves from two dominoes.

Hint 2 — first domino

Place the 4–0 domino with its 4 at row 3, column 3 and its 0 at row 3, column 4.

Hint 3 — second domino

Next, the 2–6 goes at row 1, 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: 8 must add up to 8; = must all be the same number.

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

Hard Pips answer — January 29, 2026

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

The hard board runs 10 rows by 8 columns: 16 dominoes to place across 16 regions. The mix is 7 sum regions, 6 "all equal", 3 free regions. The single cell at row 2, column 3 is marked 3, so it can only hold a 3. After that, the single cell at row 4, column 3 is marked 3, so it can only hold a 3. Rack check: 5 doubles, 5 blank halves, 2 sixes.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 2–3 goes at row 5, 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 the same number; 3 must add up to 3; 10 must add up to 10; 1 must add up to 1; 5 must add up to 5.

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

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