NYT Pips Hints & Answers for June 21, 2026

Jun 21, 2026

🚨 SPOILER WARNING

This page contains the final **answer** and the complete **solution** to today's NYT Pips puzzle. If you haven't attempted the puzzle yet and want to try solving it yourself first, now's your chance!

Click here to play today's official NYT Pips game first.

Want hints instead? Scroll down for progressive clues that won't spoil the fun.

SEE ALSO:

🎲 Today's Puzzle Overview

The easy grid, constructed by Ian Livengood, feels like a brisk warm‑up. You’ll spot a single cell with a “less than 1” constraint that can only be zero, and another cell demanding exactly a sum of 2 — both waving their answers right at you. From there, the tiny sum‑16 region clicks everything into place, letting you drop the last dominos without breaking a sweat.

Rodolfo Kurchan’s medium puzzle draws you into a checkerboard grid where only every other column is occupied. The first real foothold comes from a column of three cells forced to be equal — a rare sight in Pips — which anchors an entire flank. Pivoting between a greater‑than‑4 trigger and a pair of sum‑7 regions, you feel your way down the columns until the remaining empty cells accept their natural mates.

Kurchan’s hard grid is a masterclass in cascading equalities. Three different sum‑0 regions force zeros in far‑flung corners, and each zero unlocks a neighbor. Soon a four‑cell equals column on the right side demands all‑fives, while a three‑cell equals column nearby becomes all‑threes. The domino choices shrink so tightly that you’ll feel the puzzle solve itself, provided you trust the constraints. Today’s NYT Pips hard is a deeply satisfying logic knot.

💡 Progressive Hints

Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!

💡 Look for solo constraints
The grid has two single‑cell regions, each with a condition that practically hands you the digit. No need to guess — just scan for a cell that must be the smallest possible pip and another that tells you its exact value.
💡 Zero in on the corners
The less‑than‑1 cell sits in the third row, first column. Once you place a 0 there, its domino partner can only land next to it, and the sum‑2 cell in the second row locks that domino in place.
💡 Full layout for easy
Place the [4,2] domino vertically with 4 at row 1 col 0 and 2 at row 2 col 0. The [0,1] domino goes horizontal from row 3 col 0 (0) to row 3 col 1 (1). The [4,4] domino lands on row 0 col 1 and col 2, both 4. The [4,6] domino covers row 1 col 1 (4) and row 2 col 1 (6). Finally, the [3,5] domino places 3 at row 4 col 1 and 5 at row 4 col 0.
💡 Spot the equal column
In Kurchan’s medium, one whole column of three cells is an equals region — they must all be the same value. Find that column and think about which domino can provide two identical pips.
💡 Start on the right side
The greater‑than‑4 cell in the top right forces a 5 or 6, and it shares a sum‑7 region with the cell below it. Placing a [5,3] domino there sets off a chain through two sum‑7 regions all the way to the bottom.
💡 Full layout for medium
Place [5,3] vertically on column 4 rows 0–1 (5,3). Follow with [6,4] on column 4 rows 2–3 (4,6) to satisfy the middle sum‑7. Then [1,1] on column 4 rows 4–5 (1,1) to finish that column. The equals column forces [2,2] vertically on rows 4–5 col 0 (2,2) and then [3,2] gives 3 at row 2 col 0 and 2 at row 3 col 0. The sum‑3 region forces [0,0] on rows 0–1 col 0 (0,0). On column 2, the sum‑4 region needs [0,3] at rows 2–3 col 2 (3,0) and then [4,2] at rows 4–5 col 2 (2,4). Finally, [1,5] covers row 0 col 2 (5) and row 1 col 2 (1) to satisfy the sum‑4.
💡 Chase the zeros
Several regions in the hard puzzle demand a sum of zero — which means every cell in them must be 0. Find all the sum‑0 spots; they will be your earliest, most certain placements.
💡 Three zero anchors
You’ll find sum‑0 constraints at [3,2], [8,2], and the pair [5,0]–[5,1]. Each one forces a domino that carries a 0. Focus on those cells and their immediate domino options to anchor the left and bottom edges.
💡 The dominoes that deliver zeros
Domino [2,0] (index 14) can put 0 at [3,2] and 2 at [3,1]. Domino [0,5] (index 0) covers [8,2] with 0 and [8,1] with 5. Domino [0,6] (index 3) places 0 at [5,0] and 6 at [6,0]; the neighboring [5,1] sum‑0 then forces [0,3] to put 0 there and 3 at [5,2].
💡 Cascading equals regions
With those zeros in place, the equals region at [2,6]–[2,7] forces a domino with identical pips. The bigger equals column on column 5 (four cells) will all become the same digit — you’ll need to use the [5,5] and [3,5] dominoes to fill it. Meanwhile, the three‑cell equals at column 7 also locks into a repeated value, pulling in another domino.
💡 Full layout for hard
Place [2,0] vertically at [3,1]=2, [3,2]=0. Place [0,5] at [8,2]=0, [8,1]=5. Place [0,6] at [5,0]=0, [6,0]=6. Then [0,3] at [5,1]=0, [5,2]=3. With [6,0]=6, the equals region forces [7,0]=6 via [6,1] also giving [8,0]=1. Sum‑8 [3,0]–[3,1] gets [3,1]=2 so [3,0]=6 via [2,6] which also gives [2,0]=2. Sum‑4 [1,0]–[2,0] then needs [1,0]=2 from [2,1] (also [0,0]=1). Sum‑4 at [0,2] forces 4, so [4,5] gives [0,2]=4, [0,1]=5, completing top‑sum‑6. Sum‑8 at [1,3]–[2,3] uses [3,5] (3,5). Equals column 5: [5,5] covers [3,5]–[4,5] (5,5), [2,5] covers [2,6]=2,[2,5]=5, and [1,5] covers [6,5]=1,[5,5]=5. Equals row 2: [2,3] gives [2,7]=2,[3,7]=3. Equals column 7: [4,3] gives [4,6]=4,[4,7]=3, and [3,6] gives [5,7]=3,[6,7]=6. Finally, sum‑10 [6,3]–[7,3] uses [6,4] (4,6).

🎨 Pips Solver

Jun 21, 2026

Click a domino to place it on the board. You can also click the board, and the correct domino will appear.

✅ Final Answer & Complete Solution For Hard Level

The key to solving today's hard puzzle was identifying the placement for the critical dominoes highlighted in the starting grid. Once those were in place, the rest of the puzzle could be solved logically. See the final grid below to compare your solution.

Starting Position & Key First Steps

Pips hint for June 21, 2026 – hard level puzzle grid with critical first placements and strategy

This image shows the initial puzzle grid for the hard level, with a few critical first placements highlighted.

Final Answer: The Solved Grid for Hard Mode

NYT Pips June 21, 2026 hard puzzle full solution grid showing final answer with hints

Compare this final grid with your own solution to see the correct placement of all dominoes.

🔧 Step-by-Step Answer Walkthrough For Easy Level

1
Step 1: The sum‑2 solo cell
The region at [2,0] is a sum‑2 with only one cell, so its pip must be exactly 2. The only domino containing a 2 is [4,2]. Place it so that the 2 lands on [2,0]; the 4 must go to an adjacent cell — the only uncovered neighbor is directly above at [1,0].
2
Step 2: Less than 1 means zero
The less‑than‑1 region at [3,0] can only be 0. The sole remaining domino with a 0 is [0,1]. Put 0 on [3,0]; the 1 must go next door, which horizontally places it at [3,1] because [2,0] is already filled.
3
Step 3: The sum‑16 region demands fours
With [1,0] already a 4, the sum‑16 region now requires its other three cells ([0,1],[0,2],[1,1]) to total 12. The [4,4] domino provides two 4s, perfect for the top row’s [0,1] and [0,2]. That leaves [1,1] needing 4 to reach 12, which comes from the [4,6] domino — the 6 then lands on [2,1].
4
Step 4: The greater‑than‑4 final cell
The cell [4,0] must be greater than 4, so 5 or 6. The only unused domino is [3,5], and the 6 is already placed on [2,1]. Placing 5 at [4,0] and its partner 3 at [4,1] satisfies the unequal region (all three column 1 cells are now distinct) and completes the grid.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Trigger the greater‑than‑4 cell
The top‑right cell [0,4] is in a greater‑than‑4 region, so it must be at least 5. Domino [5,3] is the only one with a 5 or 6 that also gives a small partner. Placing it vertically with 5 at [0,4] and 3 at [1,4] immediately satisfies the constraint and starts a chain down column 4.
2
Step 2: Sum‑7 cascades down the column
The region [1,4]–[2,4] now has a 3 and needs to total 7, so [2,4] must be 4. Domino [6,4] fits, giving 6 to [3,4]. Then the next sum‑7 region [3,4]–[4,4] gets 6 and demands 1 at [4,4]; domino [1,1] supplies the 1 and also puts a 1 at [5,4], which satisfies the greater‑than‑0 constraint on that cell.
3
Step 3: The equals column locks in
The three‑cell equals region on column 0 ([3,0],[4,0],[5,0]) all share the same value. To get three equal numbers, domino [2,2] goes vertically on [4,0] and [5,0] with both cells 2, forcing [3,0] to be 2 as well. That 2 then pulls domino [3,2] into [3,0] (2) and its neighbor [2,0] (3).
4
Step 4: Sum‑3 feeds the left column
With [2,0]=3, the sum‑3 region [1,0]–[2,0] requires [1,0] to be 0. Domino [0,0] fits perfectly, covering [0,0] and [1,0] both 0. Now column 0 is fully settled.
5
Step 5: Sum‑2 and sum‑4 on column 2
The sum‑2 region [3,2]–[4,2] needs a 0‑and‑2 pair; domino [0,3] puts 0 at [3,2] and 3 at [2,2], while domino [4,2] delivers 2 at [4,2] and 4 at [5,2]. The sum‑4 region [1,2]–[2,2] then gets 1 at [1,2] from domino [1,5] (which also puts 5 at [0,2]), completing the grid.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Zeros from sum‑0 regions
Isolated sum‑0 cells [3,2] and [8,2] force zeros. Domino [2,0] (index 14) goes at [3,1]=2, [3,2]=0. Domino [0,5] (index 0) goes at [8,2]=0, [8,1]=5. The paired sum‑0 region [5,0]–[5,1] forces both cells to 0: [0,6] (index 3) gives [5,0]=0, [6,0]=6; then [0,3] (index 11) gives [5,1]=0, [5,2]=3.
2
Step 2: Equals forces the left column
The equals region [6,0]–[7,0] now has 6 at [6,0], so [7,0]=6. Domino [6,1] (index 13) fits [7,0]=6, [8,0]=1, satisfying sum‑6 [8,0]–[8,1] (1+5). The sum‑8 region [3,0]–[3,1] has [3,1]=2, so [3,0]=6; domino [2,6] (index 10) puts [2,0]=2, [3,0]=6. Sum‑4 [1,0]–[2,0] then needs [1,0]=2, so domino [2,1] (index 1) goes [1,0]=2, [0,0]=1.
3
Step 3: Top‑row sum‑6 and sum‑4
The single‑cell sum‑4 at [0,2] must be 4, so domino [4,5] (index 4) gives [0,2]=4, [0,1]=5. The top‑left sum‑6 region [0,0]–[0,1] now has 1+5, perfectly matching.
4
Step 4: Sum‑8 on column 3
The sum‑8 region [1,3]–[2,3] still needs two numbers adding to 8. The only remaining domino that sums to 8 is [3,5] (index 2). Place it vertically: 3 at [1,3], 5 at [2,3].
5
Step 5: Equals cascade on columns 5,7, and row 2
The equals region [2,5]–[3,5]–[4,5]–[5,5] forces all‑5. Place [5,5] (index 8) at [3,5]=5, [4,5]=5. Then [2,5] (index 12) goes [2,6]=2, [2,5]=5, and [1,5] (index 6) goes [6,5]=1, [5,5]=5. The equals region [2,6]–[2,7] needs both 2; with [2,6]=2, domino [2,3] (index 7) gives [2,7]=2, [3,7]=3. The equals region [3,7]–[4,7]–[5,7] forces all‑3: [4,3] (index 9) goes [4,6]=4, [4,7]=3, and [3,6] (index 5) goes [5,7]=3, [6,7]=6.
6
Step 6: Final sum‑10 and remaining cells
The sum‑10 region [6,3]–[7,3] requires 4+6. Domino [6,4] (index 15) places [7,3]=6, [6,3]=4. Every cell is now covered and all constraints are satisfied.

💡 Pro Tips for Similar Puzzles

Start with Constraints
Always begin with the most constrained regions - sum regions with small numbers or tight spaces.
Use Equal Regions
Use "equal" regions as anchors - they eliminate many possibilities quickly.
Work Systematically
Let the rules guide your placement rather than guessing randomly.
Double-Check
Verify each region's rules are satisfied before moving to the next.

🎓 Keep Learning & Improve

🧠 Strategy Guide

Master the core techniques to solve any Pips puzzle, no matter the difficulty.

🎮 How to Play

New to Pips? Learn all the rules and conditional regions in minutes.

📖 More to Read

Jun 20, 2026

NYT Pips Hints & Answers for June 20, 2026

View Answer →
Jun 19, 2026

NYT Pips Hints & Answers for June 19, 2026

View Answer →
Jun 18, 2026

NYT Pips Hints & Answers for June 18, 2026

View Answer →
Jun 17, 2026

NYT Pips Hints & Answers for June 17, 2026

View Answer →
Jun 16, 2026

NYT Pips Hints & Answers for June 16, 2026

View Answer →
Jun 15, 2026

NYT Pips Hints & Answers for June 15, 2026

View Answer →

💬 Community Discussion

Leave your comment