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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
Easy is a confidence-builder: Livengood hands you a large greater-than-15 region that funnels high-value dominos right away. Two single-cell sum‑3 anchors lock in the last of the big numbers, and a simple equals pair cleans up the rest — no branching, pure forward deductions.
Rodolfo Kurchan’s medium takes a different tack, stacking four vertical sum pairs across the top. The sum‑10 column is the pinch point: it demands a specific domino split that also feeds the adjacent sum‑7. Once that interlock clicks, the bottom row’s greater‑than and sum‑6 cage fill with whatever dominos remain.
The hard puzzle, also by Kurchan, dispenses with arithmetic almost entirely — the grid is tiled with thirty single‑cell sum targets, effectively a map of fixed numbers. The solver’s job becomes pairing the given dominos to match those pre‑determined values. It’s a methodical placement exercise that feels like a domino jigsaw, rewarding patient step‑by‑step logic.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Find the Heavy Lifter
Start by identifying which region demands the largest numbers. It’s a greater-than constraint, and it will dictate your high-value dominos.
💡 Hint 2: Pin Down the Big Cage
The greater‑than‑15 region covers cells [0,2], [0,3] and [1,2]. To exceed 15, you’ll need the 6‑pip and two 5‑pips, forcing specific dominos into that area.
💡 Hint 3: Full Placement Chain
Place the [4,6] domino with 6 in [1,2] and 4 in [2,2]. Then put [3,5] with 5 in [0,3] and 3 in [1,3]; and [5,2] with 5 in [0,2] and 2 in [0,1]. The sum‑3 cell at [1,0] takes [3,0] (3 and 0, with 0 in [0,0]). The equals pair at [2,2] and [3,2] already shares 4; the final [4,0] domino places 4 in [3,2] and 0 in [3,3] to satisfy the empty cell.
💡 Hint 1: Eye the Tightest Column
Look for the tightest vertical sum — one pair needs a total of 10, which will lock a key domino early. Focus on the top two rows.
💡 Hint 2: Split the 6‑4 Domino
The sum‑10 pair sits in column 1 ([0,1] and [1,1]), and the neighbouring sum‑7 pair is in column 2. The only way to satisfy both is to split a [6,4] domino across the two columns, with the 6 moving right into the sum‑7 cell [0,2].
💡 Hint 3: Nail the Whole Top
Place the [6,4] domino with 4 in [0,1] and 6 in [0,2]. Now sum‑10 still needs a 6 at [1,1] — supply it with the [6,2] domino, putting 2 in the empty cell [1,0]. For sum‑7, [1,2] must be 1; use [1,4] with 1 in [1,2] and 4 in [1,3]. That 4 forces [0,3]=4 from a [5,4] domino placed horizontally, which also puts 5 in [0,4] and requires a 3 in [1,4] (from the [0,3] domino, placing 0 in [1,5]). The bottom row finishes with [1,2] going to [2,1] (1) and [2,2] (2), then [4,0] covering [2,3]=4 and [2,4]=0.
💡 Hint 1: A Grid of Pre‑Set Numbers
Almost every cell carries its own sum target — therefore each cell’s value is predetermined. The puzzle is essentially a domino‑placement jigsaw.
💡 Hint 2: Row 0’s Leftmost Anchor
List the forced values. Row 0 reads 4, 4, 3, 4, 3, 3. The pair of 4s at the left can only be covered by the [4,4] domino. That’s your starting anchor.
💡 Hint 3: Finish Row 0
Next, row 0 continues: [0,2]=3 and [0,3]=4 require the [3,4] domino. The rightmost pair [0,4]=3 and [0,5]=3 must use the [3,3] domino. Now row 0 is fully placed.
💡 Hint 4: Row 1 Opens Up
Move to row 1. The forced values are [1,0]=1, [1,1]=5, [1,2]=0, [1,3]=2, [1,4]=6, [1,5]=3. The 0 only appears on the [0,5] domino, so that domino must occupy [1,2] and [1,1] (giving 0 and 5). The 1 in [1,0] needs a partner 2 in [2,0] — use the [1,2] domino vertically. The 2 in [1,3] pairs with 4 in [2,3] via [2,4] domino; the 6 in [1,4] pairs with 4 in [2,4] via [6,4] domino; and the 3 in [1,5] pairs with 1 in [2,5] via [3,1] domino.
💡 Hint 5: Complete the Lower Grid
Finish row 2 and below: [2,1]=2 must come from [3,2] domino (giving 3 to [2,2]). Row 3: [3,0]=5 takes [5,4] with 4 in [4,0]; [3,1]=2 takes [2,0] (2 and 0 to [4,1]); [3,2]=1 takes [4,1] (1 and 4 to [3,3]); [3,4]=6 takes [2,6] (6 and 2 to [3,5]). Row 4’s [4,2]=1 and [4,3]=0 get covered by [0,1] domino; [4,4]=3 and [4,5]=5 by [5,3] domino. All cells filled.
🎨 Pips Solver
May 27, 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
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
Compare this final grid with your own solution to see the correct placement of all dominoes.
🔧 Step-by-Step Answer Walkthrough For Easy Level
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Step 1: The Greater‑Than Funnel
Identify the forcing region. The greater-than-15 cage at [0,2],[0,3],[1,2] requires a sum of at least 16. With available pips, only a 6 and two 5s can manage that. So the [4,6] domino must contribute its 6, and both [3,5] and [5,2] must contribute a 5.
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Step 2: Single‑Cell Sum‑3 Anchors
The sum‑3 cells at [1,0] and [1,3] each need exactly 3. The only dominos with a 3 are [3,5] and [3,0]. Since [3,5] will use its 5 in the >15 cage, its 3 must go to [1,3]. That forces [3,0] to cover [1,0] with 3 and [0,0] with 0, obeying the less‑3 limit.
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Step 3: Drop the 6 and Final 5
Now the >15 cage sits with [0,3]=5 and [1,2] waiting for the 6. Place the [4,6] domino vertically: 6 in [1,2] and 4 in [2,2]. The sum of the cage becomes 5+6 from these two cells, so [0,2] must be at least 5. The [5,2] domino gives 5 to [0,2] and 2 to [0,1], satisfying less‑3 and pushing the total to 16.
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Step 4: Equals Pair and the Empty Cell
The equals pair at [2,2] and [3,2] must match. [2,2] already holds 4 from the [4,6] domino. The remaining [4,0] domino places 4 in [3,2] and its 0 in the empty cell [3,3], completing the grid.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The Sum‑10 Bottleneck
Concentrate on the vertical pairs. The sum‑10 column ([0,1],[1,1]) can only be made with 6+4. The domino [6,4] is the sole domino containing both 6 and 4. If you place it entirely inside sum‑10, there will be no 6 left for the neighbouring sum‑7 column, making that pair impossible. So the [6,4] domino must split across the two columns.
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Step 2: Split and Supply the 6
Place [6,4] with 4 in [0,1] and 6 in [0,2]. Now sum‑10 still lacks a 6 for [1,1]. The domino [6,2] provides that 6, giving 2 to the empty cell [1,0]. Sum‑7 now has its 6; it needs a 1 in [1,2] to total 7.
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Step 3: Feed the 1 and Cascade Right
Only the [1,4] domino carries a 1. Place it vertically: 1 in [1,2] and 4 in [1,3]. This 4 forces the sum‑8 column 3 ([0,3],[1,3]) to need a 4 in [0,3]. The [5,4] domino can supply that 4 horizontally while placing its 5 in [0,4]; this then forces [1,4] to be 3 (making 5+3=8 for sum‑8 column 4). The [0,3] domino (0‑3) fits perfectly: 3 in [1,4] and 0 in [1,5].
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Step 4: Bottom Row Greater‑Than and Sum‑6
With the top two rows done, the bottom row’s [2,1] is greater‑than‑0, while [2,2]-[2,3] sum to 6. The leftover dominos are [1,2] and [4,0]. Place [1,2] vertically: 1 in [2,1] (satisfying >0) and 2 in [2,2]. The sum‑6 pair now has 2; it needs 4 for [2,3].
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Step 5: Final Clean‑Up
The [4,0] domino fits the bill: 4 goes to [2,3] and 0 to the empty cell [2,4]. All dominos placed, puzzle solved.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Row 0’s Twin 4s
Row 0 is a straight line of single‑cell sums: 4‑4‑3‑4‑3‑3. The only way to make the adjacent pair of 4s is the [4,4] domino. Place it horizontally at [0,0]-[0,1].
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Step 2: Complete Row 0
The next two cells, [0,2]=3 and [0,3]=4, are exactly matched by the [3,4] domino; place it horizontally. The final pair [0,4]=3 and [0,5]=3 requires the [3,3] domino, also horizontal.
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Step 3: Row 1’s Hard Numbers – Zero First
Now list the forced values for row 1: [1,0]=1, [1,1]=5, [1,2]=0, [1,3]=2, [1,4]=6, [1,5]=3. The only domino that contains a 0 is [0,5]; it must occupy [1,2] (0) and [1,1] (5), satisfying the sum‑5 cell.
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Step 4: The 1-2 Start
The 1 in [1,0] needs a 2 to pair from the domino list. The [1,2] domino works perfectly: place it vertically so 1 goes to [1,0] and 2 to [2,0], meeting the sum‑2 in row 2.
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Step 5: Fill Row 1’s Middle and Link to Row 2
Move across row 1: the 2 at [1,3] demands a 4 in [2,3] (sum‑4); use the [2,4] domino vertically (2 in [1,3], 4 in [2,3]). The 6 at [1,4] pairs with 4 in [2,4] via the [6,4] domino placed vertically. The 3 at [1,5] pairs with 1 in [2,5] using the [3,1] domino vertically. Row 2 now fills: [2,0]=2, [2,1] must be 2 (sum‑2), provided by [3,2] domino with 2 in [2,1] and 3 in [2,2] (sum‑3).
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Step 6: Rows 3–4 Tiling
Complete rows 3 and 4 using the remaining dominos: [5,4] goes to [3,0]=5 and [4,0]=4 (empty); [2,0] to [3,1]=2 and [4,1]=0 (sum‑0); [4,1] to [3,2]=1 and [3,3]=4; [2,6] to [3,4]=6 and [3,5]=2 (empty); [0,1] to [4,2]=1 and [4,3]=0; [5,3] to [4,4]=3 and [4,5]=5. All cells placed.
💡 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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