NYT Pips Hints & Answers for September 5, 2026

Sep 5, 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

This Saturday’s NYT Pips set begins with a breezy Livengood easy. The solver spots two single-cell inequalities near the right edge and a couple of equal-region constraints that behave like copy commands: solve one cell and its partner immediately inherits the same pip. With a tiny domino list, the opening equals pair cascades left-to-right and finishes without much branching.

Rodolfo Kurchan’s medium turns the emphasis toward arithmetic. The solver’s eye lands on the top-side sum regions—a three-cell sum across the top and a two-cell sum just below—because they burn through the high-value dominoes quickly. Once the top row locks, a string of empty cells and a lone bottom-edge greater-than constraint complete the grid, so the difficulty lies almost entirely in that upper band.

Kurchan’s hard is the most spacious and the most anchored by tiny constraints. A single-cell sum in the lower left declares an exact value, pulling adjacent dominoes into a four-cell all-equal block. From there, the left, middle, and right edges each become small arithmetic islands—vertical sums and sum-5 pockets—so the solver can advance one forced pair at a time.

💡 Progressive Hints

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

💡 Look for the duplicate demands
Start with the equals-type regions. They force two adjacent cells to share a pip, and with only one double in the set, that tile is your natural anchor.
💡 Check the right-edge inequality
The greater-4 cell at [0,3] must hold a large pip, so its domino has to run vertically into the equal pair at [1,3]. That narrows the high-side domino immediately.
💡 Full easy answer
Place domino [3,3] horizontally at [0,1]-[0,2]. Run [2,6] vertically at [1,3]-[0,3] with 6 at [0,3] and 2 at [1,3]. Put [5,2] horizontally at [1,1]-[1,2] with 5 at [1,1], 2 at [1,2]. Place [5,4] horizontally at [2,1]-[2,0] with 5 at [2,1], 4 at [2,0]. Finish with [6,1] horizontally at [3,1]-[3,2], 6 at [3,1], 1 at [3,2].
💡 Let the sums lead
Focus on the sum-type regions first; they act like arithmetic locks. The three-cell top row and two-cell row below are far more restrictive than the empties.
💡 Open the top row
The sum region across [0,1]-[0,3] forces a very specific high split. The greater-than-0 cell at [0,0] then forces the left-edge domino into a vertical placement with the adjacent high cell.
💡 Full medium answer
Place [4,4] vertically at [0,2]-[1,2] (4 and 4). Then [0,6] goes horizontally at [1,0]-[1,1] with 0 at [1,0], 6 at [1,1]. Put [1,6] horizontally at [0,0]-[0,1] with 1 at [0,0], 6 at [0,1]. Place [0,5] vertically at [1,3]-[0,3] with 0 at [1,3], 5 at [0,3]. Put [6,6] vertically at [1,4]-[2,4] (6 and 6). Place [0,3] vertically at [2,2]-[2,3] with 0 at [2,2], 3 at [2,3]. Finish with [0,2] vertically at [2,1]-[3,1], 0 at [2,1], 2 at [3,1].
💡 Find the exact anchors
Begin with the small single-cell sum constraints and the big equals region. They reveal exact values early and anchor the surrounding dominoes.
💡 Start in the lower left
The single-cell sum at [3,0] fixes that cell and forces the adjacent domino upward into the all-equal block. That starts the left-side chain.
💡 Unlock the equal block
Once [2,0] is known from the lower-left sum, the 2x2 equals region at rows 1–2, columns 0–1 must all copy that same value. Use the [5,5] vertical in its right column and a [1,5] vertical to cover [0,0]-[1,0].
💡 Work the middle sums
The vertical sum at [1,4]-[2,4]-[3,4] resolves to high values; use [6,0] at [1,4]-[0,4] and [4,5] at [2,4]-[3,4]. Then the right-side sum-5 pair at [3,6]-[4,6] pairs with the single sum-5 at [3,7] to force [2,5] and [3,0] placements.
💡 Full hard answer
Place [3,5] vertically at [3,0]-[2,0] with 3 at [3,0], 5 at [2,0]. Put [5,5] vertically at [1,1]-[2,1] (5 and 5). Put [1,5] vertically at [0,0]-[1,0] with 1 at [0,0], 5 at [1,0]. Place [4,1] vertically at [2,2]-[3,2] with 4 at [2,2], 1 at [3,2]. Put [6,0] vertically at [1,4]-[0,4] with 6 at [1,4], 0 at [0,4]. Put [4,5] vertically at [2,4]-[3,4] with 4 at [2,4], 5 at [3,4]. Place [3,3] vertically at [3,3]-[4,3] (3 and 3). Put [6,6] vertically at [4,4]-[5,4] (6 and 6). Put [2,5] horizontally at [3,6]-[3,7] with 2 at [3,6], 5 at [3,7]. Place [3,0] vertically at [4,6]-[5,6] with 3 at [4,6], 0 at [5,6]. Put [1,2] vertically at [4,0]-[5,0] with 1 at [4,0], 2 at [5,0]. Finish with [2,6] vertically at [5,8]-[4,8] with 2 at [5,8], 6 at [4,8].

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Sep 5, 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 5, 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 5, 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: Lock the top equals pair
The two-cell equals region at [0,1] and [0,2] demands the same pip in both cells. The only domino with matching pips is [3,3], so it must sit horizontally across [0,1]-[0,2].
2
Step 2: Satisfy the greater-4 cell
The [0,3] cell is a greater-4 constraint, so it needs a pip above 4. The only remaining domino with a 6 that can enter that corner is [2,6], placed vertically with 6 at [0,3] and 2 at [1,3].
3
Step 3: Cascade down the equals pairs
With [1,3]=2, the equals region [1,2]-[1,3] forces [1,2]=2. Domino [5,2] covers [1,1]-[1,2] with 2 at [1,2], leaving [1,1]=5. That 5 then forces the vertical equals region [1,1]-[2,1] to place [5,4] at [2,1]-[2,0], giving [2,0]=4 to satisfy its greater-3 constraint.
4
Step 4: Clean up the bottom row
Remaining domino [6,1] must fill [3,1]-[3,2]. The 6 goes into the greater-5 cell at [3,1], and the 1 lands in the less-3 cell at [3,2].

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Pin the middle of the top sum
The three-cell sum across [0,1]-[0,3] is very tight. The only way to satisfy it without over/under is for [0,2] to be a 4, and the only way to place that 4 is the [4,4] domino vertically into [1,2].
2
Step 2: Fill the row below
With [1,2]=4, the two-cell sum at [1,1]-[1,2] must have [1,1]=6. The [0,6] domino then fills [1,0]-[1,1], placing 0 in the empty [1,0] and 6 in [1,1].
3
Step 3: Resolve the top-left corner
The top row still needs [0,1] and [0,3] to sum to 11. [0,0] is greater-than-0, and the only domino that can give [0,0] a positive value while supplying the top row is [1,6], so it goes horizontally at [0,0]-[0,1] with 1 and 6.
4
Step 4: Finish the top band
Now [0,3] must be 5. Place [0,5] vertically at [1,3]-[0,3] with 0 at [1,3] and 5 at [0,3]. The top band is complete.
5
Step 5: Settle the lower right
The sum region [1,4],[2,3],[2,4] still needs 15. Use [6,6] vertically at [1,4]-[2,4] for two 6s, which forces [2,3]=3. Then [0,3] covers [2,2]-[2,3] with 0 and 3. Finally, [0,2] covers [2,1]-[3,1] with 0 and 2, satisfying the greater-1 at [3,1].

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Use the single-cell sum anchor
The sum-3 constraint at [3,0] is a one-cell sum, so that cell must be exactly 3. The only domino that can occupy [3,0] with a 3 and connect upward is [3,5], so it goes vertically at [3,0]-[2,0], placing 5 at [2,0].
2
Step 2: Flood the equal block
With [2,0]=5, the four-cell equals region at rows 1-2, columns 0-1 must all be 5. The [5,5] domino covers [1,1]-[2,1] vertically, and [1,5] covers [0,0]-[1,0] vertically, placing 1 in the empty [0,0] and 5 at [1,0].
3
Step 3: Place the constrained pair below the equal block
The single-cell sum-3 at [3,0] left [3,2] as less-3 and [2,2] as greater-3. Those two adjacent cells require a low-high split, and the domino [4,1] fits exactly: vertical at [2,2]-[3,2] with 4 above and 1 below.
4
Step 4: Solve the vertical middle sum
The sum-15 column at [1,4]-[2,4]-[3,4] requires a 6 on top, 4 in the middle, and 5 at the bottom. Use [6,0] vertically at [1,4]-[0,4] (6 at [1,4], 0 at [0,4]) and [4,5] vertically at [2,4]-[3,4] (4 and 5).
5
Step 5: Build the right-side sum-5 chain
The sum-5 single at [3,7] forces 5 there, and the adjacent pair [3,6]-[4,6] sum 5. Put [2,5] horizontally at [3,6]-[3,7] for 2 and 5, then [3,0] vertically at [4,6]-[5,6] for 3 and 0.
6
Step 6: Resolve the central 15 and far right
The remaining sum-15 block [4,3]-[4,4]-[5,4] needs 3, 6, and 6, so [3,3] goes vertically at [3,3]-[4,3] (3s), [6,6] at [4,4]-[5,4], and the left-edge sum-3 uses [1,2] at [4,0]-[5,0] (1 and 2). Far right [4,8]>3 and [5,8]>1 finish with [2,6] vertical, 6 above and 2 below.

💡 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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