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NYT Pips answers — Sunday, April 5, 2026

Pips #231 · 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.

Jump to: Easy · Medium · Hard

Easy Pips answer — April 5, 2026

5×3 board · 6 dominoes · 9 regions · by Ian Livengood

The easy board runs 5 rows by 3 columns: 6 dominoes to place across 9 regions. The mix is 6 sum regions, 3 "all equal". The single cell at row 1, column 1 is marked 4, so it can only hold a 4. After that, the single cell at row 1, column 3 is marked 4, so it can only hold a 4. Rack check: 2 blank halves, 2 sixes.

Hints

Hint 1 — where to start

The single cell at row 1, column 1 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 4–6 domino with its 4 at row 1, column 1 and its 6 at row 2, column 1.

Hint 3 — second domino

Next, the 3–4 goes at row 2, column 3 → row 1, 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: 4 must add up to 4; = must all be the same number; 2 must add up to 2; 0 must add up to 0; 5 must add up to 5.

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

Medium Pips answer — April 5, 2026

3×6 board · 7 dominoes · 8 regions · by Ian Livengood

The medium board runs 3 rows by 6 columns: 7 dominoes to place across 8 regions. Rules on the board: 4 sum regions, 3 "all equal", 1 "less than". The way in: the single cell at row 2, column 1 is marked 3, so it can only hold a 3. After that, the single cell at row 2, column 6 is marked 3, so it can only hold a 3. Rack check: 1 double, 1 blank half, 2 sixes.

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

Hint 3 — second domino

Next, the 3–2 goes at row 2, column 6 → 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: = must all be the same number; 4 must add up to 4; 3 must add up to 3; <4 must add up to less than 4.

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

Hard Pips answer — April 5, 2026

5×5 board · 12 dominoes · 11 regions · by Rodolfo Kurchan

The hard board runs 5 rows by 5 columns: 12 dominoes to place across 11 regions. You're working with 9 sum regions, 2 "all equal". The way in: the single cell at row 5, column 1 is marked 3, so it can only hold a 3. After that, the two-cell region at row 1, column 5 adds to 12: both cells must be 6. Rack check: 3 doubles, 2 blank halves, 3 sixes.

Hints

Hint 1 — where to start

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

Hint 3 — second domino

Next, the 0–6 goes at row 1, column 4 → row 1, 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: 9 must add up to 9; 4 must add up to 4; = must all be the same number; 12 must add up to 12; 2 must add up to 2.

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

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