NYT Pips Hints & Answers for September 3, 2026

Sep 3, 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!

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

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🎲 Today's Puzzle Overview

Today's NYT Pips set splits neatly. Ian Livengood's easy should feel like a confidence-builder: the top-row sum regions lock the first placements, and the rest of the grid follows in a short, clean chain with no forks. Livengood's medium is a moderate step up — one small two-cell sum is the tight bottleneck, and once it resolves, the adjacent sum and equals regions unwind almost mechanically.

For medium, expect the difficulty to be front-loaded. The critical deduction sits where a tiny sum meets a row-level sum and two equals regions; solvers who spot that intersection will clear the board quickly. The bottom-left greater-than and singleton less-than constraints are mostly cleanup.

Rodolfo Kurchan's hard is a genuine jump. It has several single-cell anchors around the perimeter and a few large high-value regions, so the challenge is knowing which small constraint to pull first. Work the bottom-right corner early; otherwise the grid can look intimidating. There is little guessing here, but the chain is longer and more interdependent.

💡 Progressive Hints

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

💡 Look to the sums
Start with the sum regions, not the singleton greater-than. The two-cell sum in the top row is especially useful because it has only one way to split its total across two cells.
💡 Top row lock
The region at [0,1]/[0,2] must use the two tiles that each carry the same large pip. Placing them also fixes one cell below ([1,1]) and one cell to the right ([0,3]) because those are their tile partners.
💡 Full solve
Place the 0-5 domino vertically with 5 at [0,1] and 0 at [1,1]; place the 4-5 domino horizontally with 5 at [0,2] and 4 at [0,3]. Then put the 3-2 domino at [1,3]=3/[1,4]=2 and the 3-0 domino at [2,2]=0/[2,3]=3. Finish with the 2-6 domino at [2,0]=6/[2,1]=2.
💡 Start with the smallest sum
The key is the two-cell sum with the smallest possible total — it only has one legal pair. Find that region before worrying about the equals constraints.
💡 The [1,3]/[2,3] anchor
The two-cell sum at [1,3]/[2,3] forces a zero and a one. The zero cannot go in [0,3] because that would force the adjacent row-0 sum to use an impossible value, so the zero goes below and the three goes above.
💡 Full solve
Place the 0-3 domino vertically with 3 at [0,3] and 0 at [1,3], then the 6-1 domino with 6 at [2,2] and 1 at [2,3]. Put the 4-6 domino with 6 at [0,1] and 4 at [0,2]; the 6-6 domino vertically at [1,1]/[2,1]; the 0-0 domino vertically at [0,0]/[1,0]; the 5-5 domino vertically at [2,0]/[3,0]; and the 0-5 domino with 5 at [3,1] and 0 at [3,2].
💡 Start on the perimeter
Look for single-cell sum and greater constraints on the outer edge. They are the true anchors; the long equals region in the middle is easier after the corners are fixed.
💡 Bottom-right cluster
Begin at [8,8], [8,6], and the sum-2 pair above. The single-cell constraints at the bottom right must pair with their only adjacent cells, which drags in the sum-2 at [6,6]/[7,6] and the greater-2 at [6,7].
💡 Bottom-right detail
At [8,8] a singleton sum fixes its pip and its partner at [7,8]. Then [8,6] must be the larger half of a split whose other half at [7,6] makes the adjacent sum-2 work, forcing [6,6] to zero and [6,7] to the value that satisfies its greater-2.
💡 Top and middle
After bottom-right, use the two-cell greater-than at [0,2]/[1,2] with the double-high tile. Then move to the top-right greater-1 singleton and its adjacent equals pair; they lock the 5-3-style tile into [1,4], which starts the central equals column.
💡 Full solve
Place the 2-5 domino at [7,6]=2/[8,6]=5, the 4-1 domino at [8,8]=4/[7,8]=1, and the 3-0 domino at [6,6]=0/[6,7]=3. Put the 6-6 domino at [0,2]=6/[1,2]=6. Put the 6-3 domino at [0,5]=3/[0,6]=6 and the 5-3 domino at [0,4]=3/[1,4]=5. Then place the 4-5 domino at [2,5]=4/[2,4]=5 and the 5-5 domino at [3,4]=5/[4,4]=5. Left side: place the 6-1 domino at [3,0]=1/[4,0]=6, the 4-0 domino at [5,0]=4/[5,1]=0, the 0-2 domino at [4,2]=0/[5,2]=2, and the 1-2 domino at [2,2]=1/[3,2]=2. Finish with the 4-6 domino at [4,6]=6/[5,6]=4 and the 2-3 domino at [4,8]=2/[5,8]=3.

🎨 Pips Solver

Sep 3, 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 Sept 3, 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 Sept 3, 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 row sum locks
The two-cell sum at [0,1]/[0,2] must reach 10, which forces both cells to be 5s. The only two tiles carrying a 5 are the 0-5 and 4-5 dominoes. Place the 0-5 domino vertically with 5 at [0,1] and 0 at [1,1]; place the 4-5 domino horizontally with 5 at [0,2] and 4 at [0,3].
2
Step 2: Triple sum resolves
With [0,3] now 4, the sum-10 region at [0,3]/[1,3]/[2,3] needs the other two cells to total 6. The only way is two 3s. The 3-2 domino must put its 3 at [1,3] and its 2 at [1,4] because [1,4] is a singleton sum-2; then the 3-0 domino puts its 3 at [2,3] and 0 at [2,2].
3
Step 3: Lower sum and greater
The sum-2 region at [1,1]/[2,1]/[2,2] already has 0s at [1,1] and [2,2], so [2,1] must be 2. The final tile, the 2-6 domino, supplies that 2 at [2,1] and puts 6 at [2,0], satisfying the greater-5 singleton.
4
Step 4: Final check
The singleton sum-2 at [1,4] is already satisfied by the 3-2 domino, and every region now matches its target. No alternate orientation works after the top row is locked.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Sum of one
The sum-1 region at [1,3]/[2,3] can only be 0+1. The 0-3 domino cannot put its 0 in [0,3], because then the row-0 sum would need 7 at [0,2]. So the 0-3 domino goes 3 at [0,3] and 0 at [1,3]. The 6-1 domino then supplies 1 at [2,3] and 6 at [2,2].
2
Step 2: Row-0 sum
With [0,3]=3, the sum-7 region at [0,2]/[0,3] forces [0,2]=4. The 4-6 domino puts 4 at [0,2] and 6 at [0,1]. That satisfies part of the top-middle equals region.
3
Step 3: Equals middle
The equals region [0,1]/[1,1] now needs [1,1]=6. The 6-6 domino spans [1,1] and [2,1], and since [2,2] is already 6, the second equals region [2,1]/[2,2] is also satisfied.
4
Step 4: Top-left less
The less-5 region at [0,0]/[1,0] needs both cells below 5. The remaining 0-0 domino is the only fit; place it vertically.
5
Step 5: Bottom cleanup
The greater-1 region at [2,0]/[3,0]/[3,1] needs three cells above 1. The 5-5 domino covers [2,0]/[3,0], and the 0-5 domino covers [3,1]/[3,2] with 5 at [3,1] and 0 at [3,2], satisfying the less-3 singleton.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Bottom-right singleton anchors
The singleton sum at [8,8] must be exactly 4, so the 4-1 domino places 4 at [8,8] and 1 at [7,8], satisfying the greater-0 singleton. The singleton greater-4 at [8,6] must be 5; the 2-5 domino puts 5 at [8,6] and 2 at [7,6]. Then the sum-2 at [6,6]/[7,6] forces [6,6]=0, so the 3-0 domino places 0 at [6,6] and 3 at [6,7], satisfying greater-2.
2
Step 2: Top-middle greater-than
The two-cell greater-9 region at [0,2]/[1,2] is best locked by the 6-6 domino. Placing it vertically there gives 6 and 6, saving the double-5 for the upcoming central equals run.
3
Step 3: Top-right equals
The greater-1 singleton at [0,6] takes 6 from the 6-3 domino, putting 3 at [0,5]. The equals region [0,4]/[0,5] then demands [0,4]=3, so the 5-3 domino places 3 at [0,4] and 5 at [1,4]. This sets the first cell of the central equals column.
4
Step 4: Central equals column
Because [1,4] is 5, the entire equals region [1,4]/[2,4]/[3,4]/[4,4] must be 5s. The 4-5 domino places 5 at [2,4] and 4 at [2,5] (satisfying greater-3); the 5-5 domino covers [3,4]/[4,4] with two 5s.
5
Step 5: Left-side chain
The greater-9 pair at [4,0]/[5,0] must be 6+4. Place the 6-1 domino with 6 at [4,0] and 1 at [3,0] (greater-0). Place the 4-0 domino with 4 at [5,0] and 0 at [5,1]. Then sum-2 at [5,1]/[5,2] forces [5,2]=2, so the 0-2 domino places 2 at [5,2] and 0 at [4,2]. Finally sum-2 at [3,2]/[4,2] forces [3,2]=2, and the 1-2 domino places 2 at [3,2] and 1 at [2,2], satisfying less-3.
6
Step 6: Right-side finish
The sum-10 region at [4,6]/[5,6] is solved by the 4-6 domino with 6 at [4,6] and 4 at [5,6]. The sum-5 region at [4,8]/[5,8] is solved by the 2-3 domino with 2 at [4,8] and 3 at [5,8]. All remaining regions now match.

💡 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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