NYT Pips Hints & Answers for June 24, 2026

Jun 24, 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

Ian Livengood's easy grid presents a compact layout with five dominoes and a mix of sum, greater, and equals constraints. The deduction graph begins at a single-cell sum region, which immediately forces a specific pip value and radiates into an adjacent sum region, creating a chain that resolves the equals pairs with minimal ambiguity. This NYT Pips easy demonstrates how a lone small-sum cell can anchor an entire puzzle.

Livengood's medium puzzle expands to seven dominoes, introducing a three-cell equals block that demands uniformity. The solving architecture hinges on that equality constraint, which severely restricts pip options and pulls a low-value domino into a key position. A pair of sum-9 regions on opposite edges then leverage the forced placements, while a greater-than constraint on the left side tightens the remaining freedom. The result is a balanced graph where each deduction follows from the previous without redundancy.

Rodolfo Kurchan's hard grid is a complex web of sixteen dominoes with overlapping sum and equals regions. The most constrained points are two sum regions at the top—a sum-12 pair demanding a specific double, and a sum-4 triple that cascades into a sum-12 region below. An isolated sum-1 cell in the top-right corner acts as a second anchor, triggering a sum-9 and an equals chain down the left side, which then interlocks with a cascade of equal-value dominos along the bottom-right. The puzzle's difficulty emerges from the need to propagate constraints across disconnected clusters, requiring solvers to hold multiple partial deductions in tension.

💡 Progressive Hints

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

💡 Hint 1: Spot the isolated sum
Look for a region that contains only a single cell and has a sum target. What pip value must go there, and which dominos could supply it?
💡 Hint 2: The 5 and the chain
The sum-5 at [0,2] forces a domino with a 5. That same domino's other pip (6) will land at [1,2] and combine with the sum-10 region at [1,1]/[1,2]. The greater-5 constraint at [1,3] then demands a 6, pulling in the [6,2] domino.
💡 Hint 3: Full placement
Place [5,6] with 5 at [0,2] and 6 at [1,2]. That makes [1,1]=4 to satisfy sum-10, so place [3,4] with 4 at [1,1] and 3 at [2,1]. The greater-5 cell [1,3] takes 6 from [6,2], placing 2 at [2,3]; the equals pair forces [3,3]=2, so place [5,2] with 2 at [3,3] and 5 at [3,2]. Finally, the equals pair [2,1]/[3,1] forces [3,1]=3, so place [6,3] with 3 at [3,1] and 6 at [3,0].
💡 Hint 1: Uniformity is key
Begin with the region that forces all its cells to have the same value—an equals block of three cells. How can you achieve three identical pips with the available dominos?
💡 Hint 2: Zero cascade
The equals region at [2,2]/[3,2]/[4,2] must be all zeros because only [0,0] can give two identical low pips, and a third zero can come from a domino like [4,0] or [0,5]. Once you set that, the sum-9 at [5,1]/[5,2] forces a 5 at [5,1], and the sum-9 at [2,1]/[3,1] demands a 6 and a 3.
💡 Hint 3: Full placement
Place [0,0] at [2,2] and [3,2]. To make [4,2]=0, place [4,0] with 0 at [4,2] and 4 at [5,2]. Sum-9 at [5,1]/[5,2] then needs 5 at [5,1]—place [0,5] with 0 at [4,1] and 5 at [5,1]. For the other sum-9 at [2,1]/[3,1], place [6,2] with 6 at [2,1] and 2 at [1,1]; then [1,3] places 3 at [3,1] and 1 at [3,0]. With [1,1]=2, the greater-9 region [0,0]/[0,1]/[1,1] needs sum >9, so place [4,4] with 4 at both [0,0] and [0,1]. Finally, [5,6] places its 5 at [2,3] (greater-4) and 6 at [3,3].
💡 Hint 1: Tight sums at the top
The top-right corner holds a sum-12 pair and a single-cell sum-1. Both are extremely restrictive—focus there to find forced pip values.
💡 Hint 2: Double sixes and a lone one
The sum-12 at [0,5]/[0,6] requires a [6,6] domino. The sum-1 at [1,6] forces a 1, which must come from a [1,5] domino; its partner 5 will land at [2,6], feeding a sum-9 region.
💡 Hint 3: The sum-4 triple and its partner
The sum-4 at [0,4]/[1,4]/[2,4] can only be 2+2+0. Use [2,2] for the two 2s. The 0 must come from [0,6], which also supplies a 6 to [3,4]—that 6 pairs with another 6 from [6,4] to satisfy the sum-12 at [3,4]/[4,4].
💡 Hint 4: Left-side sum-9 and equals chain
Sum-9 at [0,0]/[0,1]/[1,0] forces three 3s. Place [3,3] at [0,0]/[0,1] and [1,3] to put 3 at [1,0] and 1 at [2,0]. That 1 starts a four-cell equals region down the left edge—use [1,1] and [4,1] to fill it with 1s, placing a 4 at [5,0] for an adjacent sum-8.
💡 Hint 5: Full placement
Top right: [6,6] at [0,5]/[0,6]; [1,5] at [1,6]/[2,6] (1,5); sum-9 at [2,6]/[3,6] gets [4,5] (4 at [3,6], 5 at [4,6]); sum-12 at [3,4]/[4,4] gets [0,6] (0 at [2,4], 6 at [3,4]) and [6,4] (6 at [4,4], 4 at [4,5]). Top left: [3,3] at [0,0]/[0,1]; [1,3] at [2,0]/[1,0] (1,3); sum-5 at [1,2]/[2,2] gets [2,3] (3,2). Left: equals needs four 1s—[1,1] at [3,0]/[3,1] and [4,1] (1,4) with 1 at [4,0] and 4 at [5,0]; sum-8 at [5,0]/[6,0] gets [0,4] (4 at [6,0], 0 at [6,1]). Bottom: greater-4 at [6,7] gets [5,3] (5,3); less-4 at [4,2]/[5,2] gets [0,1] (0,1); equals [7,7]/[8,7]/[9,7] gets [4,4] and [3,4] (4,3).

🎨 Pips Solver

Jun 24, 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 24, 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 24, 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 lone sum-5
The region at [0,2] is a single cell with a sum of 5. The only way to satisfy it is to place a domino that puts a 5 on that cell. Two dominos contain a 5: [5,6] and [5,2].
2
Step 2: Sum-10 and the 6
Placing [5,6] with 5 at [0,2] forces its other pip, 6, into the adjacent cell [1,2]. Now the sum-10 region at [1,1]/[1,2] has a 6, so [1,1] must be 4. This forces the [3,4] domino to place 4 at [1,1] and 3 at [2,1].
3
Step 3: Greater-5 and equals chain
The greater-5 constraint at [1,3] requires a pip >5, which can only be 6. The only remaining domino with a 6 is [6,2]. Place it with 6 at [1,3] and 2 at [2,3]. That 2 triggers the equals pair [2,3]/[3,3], both must be 2. Place [5,2] with 2 at [3,3] and 5 at [3,2].
4
Step 4: Final equals and empty cells
The equals pair [2,1]/[3,1] now has a 3 at [2,1] from the earlier placement, so [3,1] must also be 3. The last domino [6,3] fits perfectly, placing 3 at [3,1] and 6 at the empty cell [3,0].

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Equals block forces zeros
The three-cell equals region at [2,2]/[3,2]/[4,2] requires all cells to hold the same pip value. The only domino that provides two identical pips and works with a third matching pip is [0,0], combined with a zero from either [4,0] or [0,5]. Place [0,0] at [2,2] and [3,2].
2
Step 2: Third zero and bottom sum-9
To supply the third zero at [4,2], place [4,0] with the 0 on [4,2] and the 4 on [5,2]. Now the sum-9 at [5,1]/[5,2] has a 4 at [5,2]; to reach 9, [5,1] must be 5. Place [0,5] with 5 at [5,1] and its 0 at [4,1] (empty).
3
Step 3: Left-hand sum-9
Turn to the sum-9 at [2,1]/[3,1]. The highest available pip is 6 from [6,2]; placing it with 6 at [2,1] and 2 at [1,1] leaves [3,1] needing 3. Use [1,3] to place 3 at [3,1] and 1 at [3,0].
4
Step 4: Greater-9 on the left edge
The greater-9 region at [0,0]/[0,1]/[1,1] needs a total sum >9. With [1,1] already 2, the other two must sum to at least 8. The [4,4] domino, placing 4 at both [0,0] and [0,1], gives a total of 10, satisfying the constraint.
5
Step 5: Final greater-4 cell
The greater-4 cell at [2,3] needs a pip >4. The [5,6] domino is the only remaining domino with a pip >4 (5 or 6). Place it with 5 at [2,3] and 6 at [3,3] (empty). All cells are covered.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Top-right anchors
The sum-12 pair at [0,5]/[0,6] forces two 6s; only [6,6] can supply both. Place it there. The single-cell sum-1 at [1,6] must be a 1. Use [1,5] to place 1 at [1,6] and its partner 5 at [2,6].
2
Step 2: Sum-4 and the 0-6 domino
The sum-4 triple at [0,4]/[1,4]/[2,4] requires 2+2+0. Place [2,2] with 2 at [0,4] and [1,4]. The 0 must come from a domino with a 0 and a 6—use [0,6] to place 0 at [2,4] and 6 at [3,4].
3
Step 3: Cascading sum-9 and sum-12
The sum-9 at [2,6]/[3,6] now has a 5 at [2,6]; it needs a 4 at [3,6]. Place [4,5] with 4 at [3,6] and 5 at [4,6]. The sum-12 at [3,4]/[4,4] has 6 at [3,4] and needs another 6; place [6,4] with 6 at [4,4] and 4 at [4,5], which also satisfies the sum-9 at [4,5]/[4,6] (4+5).
4
Step 4: Top-left sum-9 and sum-5
The sum-9 at [0,0]/[0,1]/[1,0] must be 3+3+3. Place [3,3] with 3 at [0,0] and [0,1]; then [1,3] places 3 at [1,0] and 1 at [2,0]. The sum-5 at [1,2]/[2,2] needs 3+2; place [2,3] with 3 at [1,2] and 2 at [2,2].
5
Step 5: Left-side equals and sum-8
The equals region at [2,0]/[3,0]/[3,1]/[4,0] now has a 1 at [2,0] and must all be 1. Place [1,1] at [3,0]/[3,1] and [4,1] with 1 at [4,0] and 4 at [5,0]. The sum-8 at [5,0]/[6,0] uses that 4 and needs a second 4; place [0,4] with 4 at [6,0] and 0 at [6,1] (satisfying the less-3 constraint).
6
Step 6: Bottom-right constraints
Greater-4 at [6,7] takes [5,3] with 5 at [6,7] and 3 at [5,7]. Less-4 at [4,2]/[5,2] uses [0,1] with 0 at [4,2] and 1 at [5,2]. The equals chain at [7,7]/[8,7]/[9,7] gets [4,4] (4,4) at [7,7]/[8,7] and [3,4] with 4 at [9,7] and 3 at [9,8] (less-4).

💡 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 23, 2026

NYT Pips Hints & Answers for June 23, 2026

View Answer →
Jun 22, 2026

NYT Pips Hints & Answers for June 22, 2026

View Answer →
Jun 21, 2026

NYT Pips Hints & Answers for June 21, 2026

View Answer →
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 →

💬 Community Discussion

Leave your comment