NYT Pips Hints & Answers for May 24, 2026

May 24, 2026

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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!

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🎲 Today's Puzzle Overview

Ian Livengood's easy NYT Pips puzzle is anchored by a pair of rigid constraints: a single-cell sum-3 at the top-left corner and an equals region spanning the top row. These two footholds interact immediately, as the sum-3 forces a domino to deliver a 3 to [0,0], which cascades into placing a 6 in the equals region via a complementary domino. The deduction graph is linear: the equals region resolves, a sum-0 block mandates the all-zero domino, and a vertical sum-10 column integrates with a bottom sum-3 to place the final dominos in a clean, sequential chain.

Livengood's medium puzzle presents a more interleaved structure. A sum-15 row across three cells at the bottom forces a trio of 5s, immediately locking the [5,5] and [5,4] dominos. This ripples upward: the 4 in the equals region of row 2 forces the [4,1] domino, which satisfies a less-2 cell, and a sum-13 region in the same row pulls in the [1,1] and [6,6] dominos. The top region then folds in with a greater-4 constraint and an equals pair, resolved by the remaining [5,2] and [4,2] dominos. The deduction is a alternating horizontal sweep across constraints.

Rodolfo Kurchan's hard puzzle is a densely coupled system of sum-1 single-cell regions, a four-cell equals block, and multiple sum-11 zones. Numerous 1s are forced by the sum-1 cells, beginning with the [1,1] domino covering the top pair. The equals block—spanning [5,0],[5,1],[6,1],[6,2]—demands a value repeated four times, which can only be 3 given the available pips and domino counts, pulling in the [3,3], [3,6], and [3,1] dominos in a cascading sequence. The sum-11 regions then interlock with these placements, as the 6s and 5s from the domino distribution satisfy the column and pair sums. The solving graph is a web with multiple simultaneous threads, but the sum-1 anchors and the equals block form the backbone that stabilizes the rest.

💡 Progressive Hints

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

💡 Hint 1: Constraint Types
Scan for equals regions and single-cell sum constraints. These will give you early fixed values that lock down domino orientations.
💡 Hint 2: Region Focus
The equals region in the top row demands both cells match, and the isolated sum-3 at [0,0] forces a specific pip. Together they restrict which dominos can touch the top-left corner.
💡 Hint 3: Full Placement
Place the [6,3] domino at [0,0]-[0,1] (3 at [0,0], 6 at [0,1]). Place the [5,6] domino vertically at [1,2]-[0,2] (5 at [1,2], 6 at [0,2]), satisfying the equals and greater-4. The sum-0 at [1,0]-[1,1] takes the [0,0] domino. The bottom sum-3 at [5,0] uses the [3,0] domino placed vertically at [5,0]-[4,0] (3 and 0). Finally, the sum-10 column [2,0]-[4,0] gets the [5,5] domino at [2,0]-[3,0] (5+5+0=10).
💡 Hint 1: Constraint Types
Look for a sum region with a high target that dominates a row. Also note the equals and less constraints that will force precise pip matching.
💡 Hint 2: Region Focus
The sum-15 row at [3,0]-[3,2] must reach 15 with only three cells, so it likely uses 5s. Once those 5s are placed, the equals region at [2,1]-[2,2] will inherit a value and force the next domino.
💡 Hint 3: Full Placement
Place [5,5] horizontally at [3,0]-[3,1] and [5,4] vertically at [3,2]-[2,2] (5 at [3,2], 4 at [2,2]) to sum 15. The equals [2,1]-[2,2] forces [2,1]=4 via [4,1] placed horizontally at [2,1]-[2,0] (4 and 1 for less-2). Sum-13 at [2,3]-[2,5] takes [1,1] at [2,3]-[3,3] (1 and 1) and [6,6] at [2,4]-[2,5] (6+6+1=13). Top greater-4 at [1,4] gets 5 from [5,2] placed vertically at [1,4]-[1,5] (5 and 2), and equals [0,5]-[1,5] is satisfied with [4,2] at [0,4]-[0,5] (4 and 2, matching the 2 from [5,2]).
💡 Hint 1: Constraint Types
The grid is peppered with sum-1 single-cell regions. These will force many 1s. Also note the large equals block that will require a repeated pip value across four cells, and several sum-11 zones.
💡 Hint 2: First Footholds
The two adjacent sum-1 cells at [0,1] and [0,2] can be satisfied together by a single domino carrying two 1s. The equals block in the lower left ([5,0],[5,1],[6,1],[6,2]) will need a domino with a matching pair, supplemented by other dominos that deliver that same pip.
💡 Hint 3: Building the 1s and the 6
After placing the [1,1] domino on the top sum-1 pair, the single sum-1 at [1,1] gets its 1 from the [6,1] domino, which also places a 6 at [2,1] for a sum-11 region. The sum-1 at [4,0] forces a 1 there via the [3,1] domino, introducing a 3 into the equals block.
💡 Hint 4: Equals and Sum-11 Integration
The equals block becomes all 3s: the [3,3] domino covers two cells, and the [3,6] domino provides the third 3 at [6,2] while placing a 6 at [6,3] for a sum-11 region. The sum-1 at [5,3] takes the [0,1] domino. Now the sum-11 column [0,3]-[2,3] and sum-11 pair [2,1]-[2,2] can be resolved with remaining high-pip dominos.
💡 Hint 5: Full Placement
Place [1,1] horizontally at [0,1]-[0,2] (1,1). Place [6,1] vertically at [2,1]-[1,1] (6,1). Place [3,1] vertically at [5,0]-[4,0] (3,1). Place [3,3] horizontally at [5,1]-[6,1] (3,3) and [3,6] horizontally at [6,2]-[6,3] (3,6). Place [0,1] vertically at [4,3]-[5,3] (0,1). Place [2,4] vertically at [0,3]-[1,3] (2,4) and [5,5] horizontally at [2,2]-[2,3] (5,5) to satisfy the sum-11 column and pair. Place [1,5] horizontally at [7,2]-[7,3] (1,5). Place [4,1] vertically at [6,5]-[7,5] (4,1). Place [5,6] horizontally at [4,5]-[3,5] (5,6) and [2,0] horizontally at [5,5]-[5,6] (2,0) for the sum-11 quad. Place [4,4] vertically at [6,0]-[7,0] (4,4). Place [1,2] horizontally at [7,6]-[7,7] (1,2). Place [6,2] vertically at [3,6]-[3,7] (6,2) to finish the greater regions.

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May 24, 2026

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✅ 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 May 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 May 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: Top-Left Anchor
The single-cell sum-3 at [0,0] forces that cell to be 3. The only available dominos containing a 3 are [6,3] and [3,0]. Since [0,0] must pair with an adjacent cell, the [6,3] domino fits at [0,0]-[0,1], placing 3 at [0,0] and 6 at [0,1].
2
Step 2: Equals Completion
The equals region at [0,1]-[0,2] now has [0,1]=6, so [0,2] must also be 6. The domino [5,6] placed vertically at [1,2]-[0,2] gives [0,2]=6 and [1,2]=5, satisfying the adjacent greater-4 region (5 > 4).
3
Step 3: Sum-0 Forces Zero
The sum-0 region at [1,0]-[1,1] requires both cells to be 0. The only domino with two zeros is [0,0]. Place it horizontally covering [1,0] and [1,1].
4
Step 4: Column Chain
The sum-3 region at [5,0] demands a 3. Place the [3,0] domino vertically at [5,0]-[4,0], giving [5,0]=3 and [4,0]=0. With [4,0]=0, the sum-10 column [2,0]-[4,0] needs 10 from the remaining two cells, so place the [5,5] domino horizontally at [2,0]-[3,0] for 5+5+0=10.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Sum-15 Row
The three cells in row 3, columns 0–2 must sum to 15. The maximum per cell is 6, so the only feasible combination is 5+5+5. The [5,5] domino covers two cells, placed horizontally at [3,0]-[3,1] (both 5). The third cell [3,2] also needs 5, so place the [5,4] domino vertically with [3,2]=5 and [2,2]=4.
2
Step 2: Equals and Less-2
The equals region [2,1]-[2,2] now has [2,2]=4, forcing [2,1]=4. The [4,1] domino placed horizontally at [2,1]-[2,0] gives [2,1]=4 and [2,0]=1, which satisfies the less-2 constraint (1 < 2).
3
Step 3: Sum-13 Region
The sum-13 region spans [2,3],[2,4],[2,5]. With [2,3] able to pair with the empty cell [3,3], place the [1,1] domino vertically at [2,3]-[3,3] to put 1 at [2,3]. The remaining sum of 12 must come from [2,4] and [2,5], so place the [6,6] domino horizontally at [2,4]-[2,5] for 6+6+1=13.
4
Step 4: Greater-4 and Top Equals
The greater-4 region at [1,4] requires a pip >4. The remaining dominos are [4,2] and [5,2]. [5,2] can provide a 5, so place it vertically at [1,4]-[1,5] giving [1,4]=5 and [1,5]=2. The equals region [0,5]-[1,5] must match: with [1,5]=2, [0,5] must be 2. Place [4,2] vertically at [0,4]-[0,5] ([0,4]=4 empty, [0,5]=2), completing the grid.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Top Sum-1 Pair
The sum-1 cells at [0,1] and [0,2] are adjacent and both must be 1. The only domino with two 1s is [1,1], so place it horizontally at [0,1]-[0,2].
2
Step 2: Single Sum-1 and a 6
The sum-1 at [1,1] needs a 1. The [6,1] domino can place a 1 there by orienting vertically at [2,1]-[1,1], giving [1,1]=1 and [2,1]=6. This 6 will help satisfy the sum-11 pair [2,1]-[2,2].
3
Step 3: Lower Sum-1 Seeds Equals
The sum-1 at [4,0] forces a 1. Place the [3,1] domino vertically at [5,0]-[4,0] to set [4,0]=1 and [5,0]=3. This inserts the first 3 into the four-cell equals block.
4
Step 4: Complete Equals Block
The equals block [5,0],[5,1],[6,1],[6,2] must all be 3. With [5,0]=3, place [3,3] horizontally at [5,1]-[6,1] (both 3). For the last cell [6,2], place [3,6] horizontally at [6,2]-[6,3] (3 and 6). The sum-1 at [5,3] then takes [0,1] vertically at [4,3]-[5,3] (0 and 1).
5
Step 5: Sum-11 Column and Pair
The sum-11 column [0,3]-[2,3] requires totals. Place [2,4] vertically at [0,3]-[1,3] (2 and 4). To reach 11, [2,3] needs 5, so place [5,5] horizontally at [2,2]-[2,3] (5,5). This also satisfies the sum-11 pair [2,1]-[2,2] because [2,1]=6 and [2,2]=5 sum to 11.
6
Step 6: Remaining Sum-1s and Regions
The sum-1 at [7,2] is covered by [1,5] horizontally at [7,2]-[7,3] (1 and 5); this 5 plus [6,3]=6 completes the sum-11 [6,3]-[7,3]. The sum-1 at [7,5] uses [4,1] vertically at [6,5]-[7,5] (4,1). Place [5,6] horizontally at [4,5]-[3,5] (5,6) for the greater region, then [2,0] horizontally at [5,5]-[5,6] (2,0) to finish the sum-11 quad (5+2+0+4=11, with [6,5]=4). Place [4,4] vertically at [6,0]-[7,0] (4,4) for the equals pair. Finally, [1,2] at [7,6]-[7,7] (1,2) for the sum-1 and greater-1, and [6,2] vertically at [3,6]-[3,7] (6,2) to close the greater regions.

💡 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.

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