NYT Pips Hints & Answers for August 24, 2026

Aug 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!

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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

Ian Livengood's easy grid opens as a small deduction graph with two independent footholds. A maximal two-cell sum region in the upper right collapses to a single double-domino candidate, while the central equals region shapes the middle placements. The left-side low-sum pair then closes the pips without branching.

Rodolfo Kurchan's medium grid is structured around a pair of top-row equals regions. Because each contains a two-cell horizontal pocket, two double-domino anchors are forced early; those placements then transfer their values down through adjacent vertical cells. From there the grid flows left to right through a two-cell sum, two single-cell less-than constraints, and a right-side equals-sum chain.

Kurchan's hard grid is more heavily interwoven. Here the key top-row equals region has no natural double, so its matching pips must enter from two different dominoes, one from the left and one from the right, producing a top-row cascade that propagates into a sum region and a column-shaped low-sum region. The deduction graph then exploits three different high pips, a vertical greater-than region, and a cluster of tiny sums to force the remaining placements in a tight sequence. This NYT Pips hard rewards recognizing how split equals regions transmit constraints to their neighbors.

💡 Progressive Hints

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

💡 Scan the sums
Look for a two-cell sum region whose target is so extreme that it can only be built from the highest available pips; that is your first anchor.
💡 Isolate the right anchor
The horizontal two-cell sum at row 1, columns 3–4 is the tightest endpoint. Solving it frees the only double-high domino and forces the vertical sum-12 group to its left to take its high pips from two different dominoes.
💡 Lock the full chain
Place [6,6] horizontally at [1,3]–[1,4]. The L-shaped equals region forces [2,2] horizontally at [2,2]–[2,3], so [1,2] must be 2; place [6,2] horizontally at [1,1]–[1,2] with 6 at [1,1]. Use [6,1] vertically at [2,1]–[2,0] with 1 at [2,0], then finish [1,0] vertically at [0,0]–[1,0] with 1 above 0.
💡 Equality first
Look for equals regions with two cells in the same row; a double-domino candidate can often lock the entire region at once.
💡 Anchor the top row
The two top-row equals regions at row 0 columns 1–2 and row 0 columns 3–4 are the key. Each one will commandeer a double, leaving the third cell of each region to transfer the forced value downward.
💡 Finish the cascade
Place [4,4] horizontally at [0,1]–[0,2]; then [1,1] must be 4, so [4,3] goes vertical at [1,1]–[1,0] with 4 above 3. Place [3,3] horizontally at [0,3]–[0,4]; then [1,4] must be 3, so [3,2] goes vertical at [1,4]–[2,4] with 3 above 2. The sum-4 region takes [2,2] horizontally at [1,2]–[1,3]. The less-2 cell [1,5] takes [1,5] vertical with [2,5], forcing [2,6] to 5 via [5,3]; finish [6,4] horizontal at [3,4]–[3,5] with 6 over 4.
💡 Spot the split equals
Look for an equals region that cannot be satisfied by a double domino; it must drag its required pips in from two separate dominoes.
💡 Start at top row
Start at row 0, columns 2–3. Placing those equal 4s forces the two neighboring top-row cells to be 5 and 3, which then sets the top-left sum-10 in motion.
💡 Drop through the left column
The top-left sum-10 now forces [0,0] to 5, so [1,0] becomes 0 from the [5,0] domino; that activates the sum-1 column and forces [2,0] to 1 via [6,1], placing a 6 at [3,0].
💡 Chain the middle
The other greater-8 cell at [4,0] then must be 6, so use [6,2] vertically with 2 at [5,0]. Next, the big four-cell equals region locks [1,4]–[2,4] as [3,3], and the adjacent sum-2 forces [1,2]–[1,3] via [3,0], setting off the middle double-2.
💡 Full final chain
Place [4,5] horizontally at [0,1]–[0,2] (5,4) and [3,4] horizontally at [0,3]–[0,4] (4,3). Next, [5,0] goes vertical at [0,0]–[1,0] (5,0); then [6,1] goes vertical at [2,0]–[3,0] (1,6). Place [6,2] vertical at [4,0]–[5,0] (6,2). In the middle, place [3,3] vertical at [1,4]–[2,4], [3,0] horizontal at [1,2]–[1,3] (0,3), and [2,2] horizontal at [2,2]–[2,3]. Finish with [0,6] vertical at [3,4]–[4,4] (6,0), [1,4] vertical at [3,3]–[4,3] (4,1), [0,1] vertical at [5,3]–[5,4] (1,0), and [5,5] horizontal at [5,1]–[5,2].

🎨 Pips Solver

Aug 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 August 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 August 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: Max out the right sum
The horizontal two-cell region at [1,3]–[1,4] must reach the highest possible total. Since [1,4] has no other neighbor, the only viable covering is the [6,6] domino placed horizontally.
2
Step 2: Resolve the equals L
The L-shaped equals region cannot put the double-2 domino under [1,2] without stranding [2,3], so [2,2] must cover [2,2]–[2,3] horizontally. With [1,3] already occupied, [1,2] must take its matching 2 from the [6,2] domino, placed horizontally at [1,1]–[1,2].
3
Step 3: Feed the left sum
The vertical sum-12 region at [1,1]–[2,1] now has [1,1]=6, so [2,1] must also get 6. The remaining [6,1] domino goes vertically at [2,1]–[2,0], placing its 1 in [2,0].
4
Step 4: Close the low sum
With [2,0]=1, the adjacent sum-1 region forces [1,0] to be 0. This leaves the [1,0] domino to run vertically at [0,0]–[1,0] with 1 above 0.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Anchor top-left equals
The region [0,1],[0,2],[1,1] must be equal. Its horizontal pair [0,1]–[0,2] is best anchored by the [4,4] domino; place it there. Then [1,1] must also be 4, and the only adjacent low cell is [1,0], so [4,3] goes vertical with 4 at [1,1] and 3 at [1,0].
2
Step 2: Anchor top-right equals
The top-right equals region [0,3],[0,4],[1,4] must also match. Place [3,3] horizontally at [0,3]–[0,4], making [1,4] equal 3. The [3,2] domino then goes vertical at [1,4]–[2,4] with 3 above 2.
3
Step 3: Fill the central sum
The two-cell sum-4 region at [1,2]–[1,3] cannot share pips with the adjacent equals regions, so the [2,2] domino is forced to sit horizontally there.
4
Step 4: Open the right side
The singleton less-2 cell at [1,5] must be 1. Place [1,5] vertically at [1,5]–[2,5] with 1 above 5; this forces its equal neighbor [2,6] to 5, so [5,3] goes vertical at [2,6]–[3,6] with 5 above 3.
5
Step 5: Complete the final sum
The sum-8 region [2,4]–[3,4] already has 2 at [2,4], so [3,4] must be 6. Place [6,4] horizontally at [3,4]–[3,5] with 6 at [3,4] and 4 at [3,5], satisfying the sum-7 region.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Split the top equals
The equals region at [0,2],[0,3] must produce matching 4s, but there is no double-4. Therefore the two 4s must come from [4,5] and [3,4]: place [4,5] horizontally at [0,1]–[0,2] (5,4) and [3,4] horizontally at [0,3]–[0,4] (4,3).
2
Step 2: Cascade into the left column
With [0,1]=5, the top sum-10 region forces [0,0]=5. The only way to cover [0,0] without breaking the nearby sums is [5,0] vertical at [0,0]–[1,0] (5,0). Then sum-1 forces [2,0]=1, so [6,1] goes vertical at [2,0]–[3,0] (1,6), satisfying the greater-8 cell at [3,0].
3
Step 3: Supply the second greater-8
The other cell in the greater-8 region, [4,0], must be 6. Place [6,2] vertical at [4,0]–[5,0] with 6 at [4,0] and 2 at [5,0], which also satisfies the singleton sum-2 region.
4
Step 4: Lock the big equals
The four-cell equals region already has [0,4]=3. Place [3,3] vertically at [1,4]–[2,4] (3,3) to extend it. Then [1,3] must also be 3. Since [0,3] and [1,4] are already covered, and [2,3] must remain available for the double-2, it takes the [3,0] domino horizontally at [1,2]–[1,3] with 0 at [1,2] and 3 at [1,3].
5
Step 5: Resolve the central sums
The sum-2 region [1,2]–[2,2] now has [1,2]=0, so [2,2] must be 2. Place [2,2] horizontally at [2,2]–[2,3] with 2 at both cells, satisfying the empty cell.
6
Step 6: Finish the bottom
The sum-10 at [3,3]–[3,4] needs 4 and 6. Place [0,6] vertical at [3,4]–[4,4] (6,0), then [1,4] vertical at [3,3]–[4,3] (4,1). The bottom sum-2 then forces [5,3]=1, so [0,1] goes vertical at [5,3]–[5,4] (1,0). Finally, the equals region at [5,1]–[5,2] is completed by [5,5] horizontally.

💡 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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