NYT Pips Hints & Answers for August 23, 2026

Aug 23, 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

Easy is a confidence-builder — a compact three-row grid built around two equals pairs, a single-cell greater-than, a less-than, and a lone sum cell. Livengood keeps the forks minimal; once you see which equals pair anchors the left column, the rest of the solve is a straight shot. Expect quick, clean deductions.

Medium is Rodolfo Kurchan's tight-wire act. The grid has one tall three-cell equals corridor and two smaller equals pairs, plus a far-right sum pair. The bottleneck is that three-cell corridor: it forces one of the only repeated-number dominos into place and drags a second domino into the bottom row. Find that latch and the top-right follows.

Hard is Rodolfo Kurchan's sum-architecture grind. Nearly every region is a sum, and two three-cell towers dominate the solving. The bottleneck is arithmetic rather than parsing: one tower demands a repeated high number plus a low value, while the opposite tower demands a repeated mid number plus another low value. This NYT Pips hard will feel slow and methodical rather than flashy.

💡 Progressive Hints

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

💡 Start with equalities
Look for the two equals regions first—they force matching pip values across adjacent cells. The single-cell greater-than, less-than, and sum regions will then narrow those matches quickly.
💡 Anchor the left column
The equals region at [0,1]/[1,1] is the key. It forces both cells to the same high value; the only way to satisfy the sum target at [1,0] while placing a matching value at [1,1] is to use the domino with [4,6] vertically. The top-left becomes a clean chain.
💡 Full easy solution
Place [5,6] horizontally so [0,1]=6 and [0,2]=5. Place [4,6] vertically so [1,1]=6 and [1,0]=4; that satisfies the sum at [1,0]. Place [2,3] horizontally at [0,3]=3 and [0,4]=2, then [5,2] vertically at [1,4]=2 and [1,5]=5. Finish with [0,6] horizontally at [2,2]=0 and [2,3]=6.
💡 Follow the equals regions
The medium hinges on equals regions—including one tall three-cell band. Treat the far-right sum pair and the greater-than cell as boundaries that eliminate candidate pip values.
💡 Use the three-cell corridor
The three-cell equals column at [1,1]/[2,1]/[3,1] is the bottleneck. It must use the [3,3] domino to cover two cells, and the leftover [0,3] domino to supply the third matching cell from the bottom row. That forces the bottom sum pair.
💡 Full medium solution
Place [1,1] vertically at [1,0]/[2,0] for the left equals pair. Place [3,3] vertically at [1,1]/[2,1], and put [0,3] horizontally at [3,1]=3, [3,2]=0. Use [4,5] vertically at [1,2]=5, [2,2]=4; then [0,5] vertically at [1,3]=5, [2,3]=0. Put [2,5] horizontally at [0,4]=5, [0,5]=2, and [2,0] horizontally at [1,4]=0, [1,5]=2.
💡 Read the sum map
This puzzle is nearly all sum regions. Start by separating the three-cell sums from the two-cell and single-cell sums—the three-cell towers are the structural anchors.
💡 Find the twin towers
Focus on the vertical three-cell sum at [4,0]/[5,0]/[6,0] and its right-side twin at [4,3]/[5,3]/[6,3]. Each tower forces a repeated-number domino; the left one also forces a low-value domino, and the right one forces a low-value domino paired with a high value.
💡 Lock the left tower
The left tower [4,0]/[5,0]/[6,0] must be a repeated high number plus a low value. That means the [6,6] domino goes vertically in the lower two cells, and [0,2] must cover [4,0] with the low value at [4,0] and a 2 at [3,0].
💡 Resolve the right tower
The right tower [4,3]/[5,3]/[6,3] needs a repeated mid number plus a low value. Place [4,4] vertically at [5,3]/[6,3], then [6,1] horizontally at [4,2]/[4,3] so [4,3] gets the low value. That also completes the mid-left three-cell sum with [0,6] at [3,1]/[3,2].
💡 Full hard solution
Put [6,6] vertically at [5,0]/[6,0] and [0,2] vertically at [3,0]=2, [4,0]=0 to finish the left tower. Put [4,4] vertically at [5,3]/[6,3], and [6,1] horizontally at [4,2]=6, [4,3]=1. Put [0,6] horizontally at [3,1]=0, [3,2]=6. Now top-left: [1,3] vertical at [1,0]=3, [2,0]=1; [5,0] horizontal at [0,0]=5, [0,1]=0; [6,4] horizontal at [0,2]=4, [0,3]=6. Then [5,1] vertical at [1,3]=1, [2,3]=5. Finish right side: [4,3] horizontal at [4,5]=4, [4,6]=3; [4,2] vertical at [5,5]=4, [6,5]=2.

🎨 Pips Solver

Aug 23, 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 August 23, 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 August 23, 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: Read the left-side lock
The equals region at [0,1]/[1,1] forces both cells to the same value. Since [0,2] is a greater-than-4 cell, its value must be 5 or 6. The domino [5,6] is the only one that can put a 6 at [0,1] and a 5 at [0,2], so it locks horizontally in row 0.
2
Step 2: Complete the first equals pair
With [0,1]=6, [1,1] must also be 6. The sum region at [1,0] must total 4, so [1,0] has to be 4. Place [4,6] vertically with 4 at [1,0] and 6 at [1,1].
3
Step 3: Solve the top-right equals
The less-than-4 cell at [0,3] can be 3 or lower, and [0,4]/[1,4] must match. The [2,3] domino at [0,3]=3, [0,4]=2 and [5,2] at [1,4]=2, [1,5]=5 satisfy both the less-than and equals constraints.
4
Step 4: Finish the bottom row
Remaining cells [2,2] and [2,3] take [0,6] horizontally. The greater-than-5 region at [2,3] needs a 6, and [2,2] takes the 0.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Use the tall equals corridor
The region at [1,1]/[2,1]/[3,1] must have three identical values. The only domino that can supply two equal values for that number is [3,3], so it must stand vertically at [1,1]/[2,1]. The third matching cell at [3,1] must come from another domino with a 3—[0,3].
2
Step 2: Force the bottom sum pair
Place [0,3] horizontally at [3,1]=3 and [3,2]=0. Since [2,2]/[3,2] sum to 4, [2,2] must be 4, so [4,5] goes vertically at [1,2]=5 and [2,2]=4.
3
Step 3: Complete the horizontal equals
The region [1,2]/[1,3] now forces [1,3] to also be 5. That pulls [0,5] vertically into [1,3]=5 and [2,3]=0, satisfying the empty cell at [2,3].
4
Step 4: Lock the left equals pair
With the middle solved, the remaining equals region [1,0]/[2,0] needs two matching low values. Place [1,1] vertically at [1,0]=1 and [2,0]=1.
5
Step 5: Finish the top-right sum
The sum-4 region at [0,5]/[1,5] now requires a pair of 2s. Place [2,5] horizontally at [0,4]=5 and [0,5]=2; then [2,0] horizontally at [1,4]=0 and [1,5]=2. The greater-than-1 cell at [0,4] now checks out.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Anchor the left tower
The three-cell sum at [4,0]/[5,0]/[6,0] must total 12. The only viable structure is two 6s and a 0, so place [6,6] vertically at [5,0]/[6,0].
2
Step 2: Complete the left tower
[4,0] must now be 0, and the remaining domino with 0 and 2 fits [3,0]/[4,0] vertically: [3,0]=2, [4,0]=0. That also makes the two-cell sum at [2,0]/[3,0] total 3, forcing [2,0]=1.
3
Step 3: Build the right tower
The three-cell sum at [4,3]/[5,3]/[6,3] must total 9, which needs 1+4+4. Place [4,4] vertically at [5,3]/[6,3], then [6,1] horizontally at [4,2]=6 and [4,3]=1.
4
Step 4: Fill the central sum
The horizontal three-cell sum at [3,1]/[3,2]/[4,2] needs 12 and is now pinned: [4,2]=6, so the remaining two must be 0 and 6. Place [0,6] horizontally at [3,1]=0, [3,2]=6.
5
Step 5: Resolve the upper-left sums
From [2,0]=1 and [3,0]=2, the sum at [1,0]/[2,0] gives [1,0]=3 via [1,3] vertical at [1,0]/[2,0]. Then the sum at [0,0]/[1,0] forces [0,0]=5, and [0,1]/[0,2] sum to 4 with 0 and 4: place [5,0] horizontally at [0,0]/[0,1] and [6,4] horizontally at [0,2]/[0,3].
6
Step 6: Finish right and bottom
The sum-7 region at [0,3]/[1,3] now has [0,3]=6, so [1,3]=1 from [5,1] vertical at [1,3]/[2,3], satisfying the single sum-5 at [2,3]=5. Finally, place [4,3] horizontally at [4,5]=4/[4,6]=3 and [4,2] vertically at [5,5]=4/[6,5]=2 to complete the equals pair, greater-than cell, and sum-2.

💡 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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