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NYT Pips answers — Wednesday, January 28, 2026

Pips #164 · 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 — January 28, 2026

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

Easy today is a 4×5 board with 5 dominoes and 6 regions. Rules on the board: 3 sum regions, 2 "all equal", 1 free region. The way in: the single cell at row 2, column 5 is marked 4, so it can only hold a 4. After that, the "=" pair at row 2, column 3 needs two identical pips — a double, or matching halves from two dominoes.. Rack check: 2 doubles, 1 blank half.

Hints

Hint 1 — where to start

The single cell at row 2, 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 4–1 domino with its 4 at row 2, column 5 and its 1 at row 3, column 5.

Hint 3 — second domino

Next, the 5–5 goes at row 2, column 2 → row 2, 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: 5 must add up to 5; = must all be the same number; 4 must add up to 4; 8 must add up to 8.

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

Medium Pips answer — January 28, 2026

4×6 board · 7 dominoes · 10 regions · by Ian Livengood

7 dominoes, 10 regions, a 4×6 grid — that's the medium puzzle for January 28, 2026. Rules on the board: 4 sum regions, 2 "all equal", 2 "greater than", 2 free regions. Best foothold — the single cell at row 1, column 1 is marked 3, so it can only hold a 3. After that, the single cell at row 1, column 6 is marked 3, so it can only hold a 3. Rack check: 2 doubles, 1 blank half, 3 sixes.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 5–3 goes at row 1, column 5 → row 1, 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: 3 must add up to 3; = must all be the same number; 9 must add up to 9; >2 must add up to more than 2; >5 must add up to more than 5.

3
=
3
9
>2
>5
4
=
Solution as text
  1. 5–2 — 5 at row 3, column 1, 2 at row 3, column 2
  2. 3–4 — 3 at row 1, column 1, 4 at row 2, column 1
  3. 5–3 — 5 at row 1, column 5, 3 at row 1, column 6
  4. 0–6 — 0 at row 3, column 6, 6 at row 2, column 6
  5. 4–5 — 4 at row 3, column 5, 5 at row 2, column 5
  6. 6–6 — 6 at row 1, column 2, 6 at row 2, column 2
  7. 1–1 — 1 at row 4, column 3, 1 at row 4, column 4

Hard Pips answer — January 28, 2026

7×5 board · 10 dominoes · 11 regions · by Rodolfo Kurchan

The hard board runs 7 rows by 5 columns: 10 dominoes to place across 11 regions. The mix is 4 sum regions, 1 "all equal", 1 "less than", 1 "greater than", 3 free regions. The single cell at row 1, column 1 is marked 3, so it can only hold a 3. After that, two cells adding to more than 10 at row 3, column 5: that's 5+6 or 6+6 territory.. Rack check: 3 doubles, 4 blank halves, 1 six.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 2–6 goes at row 2, column 5 → row 3, column 5. 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 3; 5 must add up to 5; = must all be the same number; 2 must add up to 2; >10 must add up to more than 10.

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

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