NYT Pips Hints & Answers for July 20, 2026

Jul 20, 2026

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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

Ian Livengood's easy grid opens on two independent footholds: a sum-10 region that mandates a pair of 5s, and a less-1 cell demanding a zero. These lock the two 5-bearing dominoes into a vertical column, which in turn fixes a 2 via a greater-1 constraint, cascading to a sum-4 region and an equals pair. The remaining placements fall neatly from the available domino pool.

Rodolfo Kurchan's medium puzzle anchors on a sum-0 region—the only way to satisfy zero is with two zeros—forcing both cells in that region to zero and splitting the zero-bearing dominoes across the grid. This radiates into a sum-5 row, an equals column, and a greater-3 cell that demands a 4 or higher, ultimately resolving through a chain of linked dominoes where each placement resolves the next.

Kurchan's hard grid is a dense tapestry of equals regions that act as constraint multipliers. A massive four-cell equals block compels all cells to the same pip value, which from the domino list can only be 6, consuming every domino with a 6. Parallel equals clusters for 3s, 2s, and 1s interlock with sum targets—especially a sum-1 region that forces a 1 adjacent to the 2-equals block—threading a narrow solution path through the domino set. This NYT Pips hard demands tracing equalities across multiple dominoes simultaneously.

💡 Progressive Hints

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

💡 Hint 1: Spot the Extreme Sum
Focus on the region with a target sum that is unusually high for the available dominoes. The pip values available restrict the possible pairs that can achieve that total.
💡 Hint 2: Zero In on the 5s
The sum-10 region occupies two vertically adjacent cells in the left column area. Two different dominoes in the set each contain a 5—consider how they can be stacked to reach exactly 10.
💡 Hint 3: Full Solution Path
Place the [0,5] domino with 0 in the less-1 cell [1,0] and 5 in [1,1]. Place [2,5] with 5 in [2,1] and 2 in [2,0] to satisfy greater-1. The equals pair [0,3]/[1,3] both need 2s: use [4,2] with 2 in [1,3] and 4 in [2,3], leaving [3,3] needing 0 from [0,3] (with 3 in [3,2]). Finally, [2,6] fills [0,2] with 6 and [0,3] with 2, completing the grid.
💡 Hint 1: The Zero-Sum Trap
Locate the region where the target sum forces a unique digit in multiple cells. With pips ranging from 0 to 6, only one combination can add to this particular total.
💡 Hint 2: Split the Zeros
The sum-0 region is a horizontal pair near the middle of the grid. Because only 0+0 equals 0, both cells must be 0, requiring two distinct dominoes that carry a 0 pip to cover them.
💡 Hint 3: Full Solution Path
Place [2,0] with 2 in [3,0] and 0 in [2,0], and [5,0] with 5 in [2,2] (greater-3) and 0 in [2,1], satisfying sum-0. The sum-5 at [3,0]/[3,1] now has 2 and needs 3: use [3,1] to put 3 in [3,1] and 1 in [3,2]. The equals column [3,2]/[4,2] forces [0,1] to place 1 in [4,2] and 0 in [4,1]. Then [1,2] gives 1 to less-2 [0,0] and 2 to [1,0]; [3,4] gives 3 to [1,1] and 4 to [1,2] (>3). Finally, [0,0] fills the equals [2,3]/[3,3] with 0,0.
💡 Hint 1: Find the Monolith
Look for the largest equals region on the board. Its size dictates that only one pip value can appear in enough dominoes to occupy all its cells.
💡 Hint 2: The 2x2 Block of Sixes
The 2x2 equals block centered around rows 3–4 and columns 2–3 requires four cells with the same digit. Only pip 6 appears in enough dominoes (six total mentions) to cover four distinct cells, so this whole block must be 6.
💡 Hint 3: A Ten with a Six
Adjacent to the 6-block, a sum-10 region at [4,0]/[5,0] needs 4+6. Since most 6s are already consumed by the block, one remaining 6 must pair with a 4 to satisfy this region—pointing directly to the [4,6] domino.
💡 Hint 4: The Triple-2 Cascade
A three-cell equals region at [0,4]/[1,4]/[1,5] forces all to be 2. This links to the pair of 5s at [0,2]/[0,3] through the [5,2] domino, which can place a 2 in [0,4] and a 5 in [0,3] in one move.
💡 Hint 5: Full Solution Summary
Place [5,2] with 5 at [0,3] and 2 at [0,4]—that instantly fixes the 5-equals and feeds the 2-equals. Use [2,2] to put 2s at [1,4]/[1,5]. The sum-1 [1,2] forces [1,5] domino to put 1 there and 5 at [0,2]. Sum-0 row (2,2–2,4) gets 0s from [0,0] at [2,3]/[2,4] and [0,6] at [2,2]/[3,2] (6 goes into block). The 6-block deploys [6,6] at [6,3]/[7,3], [1,6] at [3,3]/[3,4], and [6,3] at [4,3]/[5,3]. The sum-10 at [4,0]/[5,0] uses [4,6] (4 and 6), and greater-3 [4,1] takes 5 from [5,6] (5,6) with 6 into block. Three 3s in equals [5,2]/[5,3]/[6,2] come from [3,3] at [5,2]/[6,2] and [6,3] at [5,3]. Finally, [2,6] fills [7,2]/[8,2] (6 and sum-2's 2), [5,5] fills [7,4]/[8,4] with 5s, and [1,1] plus [0,1] complete the equals-1 block.

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Jul 20, 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 July 20, 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 July 20, 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 Sum-10
The sum-10 region at [1,1]/[2,1] must be 5+5 because the available dominoes include two 5s (in [0,5] and [2,5]) and no other pair sums to 10. Both cells are forced to 5, demanding that the two 5-bearing dominoes occupy these vertically adjacent cells.
2
Step 2: Zero and the First 5
The less-1 cell [1,0] demands a 0. The only domino that can supply a 0 while placing a 5 in the sum-10 region is [0,5]. Place it with 0 at [1,0] and 5 at [1,1]. Then the other 5 must go immediately below: place [2,5] with 5 at [2,1] and its 2 at [2,0], which conveniently satisfies the greater-1 constraint on [2,0].
3
Step 3: Equals and the Sum-4
The equals region [0,3]/[1,3] requires matching values. The remaining domino [4,2] contains a 2, so place it with 2 at [1,3] and 4 at [2,3]. The sum-4 region [2,3]/[3,3] now reads 4 + ? = 4, forcing [3,3] to be 0. The only remaining domino with a 0 is [0,3]; place its 0 at [3,3] and its 3 at the empty cell [3,2].
4
Step 4: The Final Corner
Cell [0,2] lies in a greater-5 region. The last unused domino [2,6] supplies the needed 6 at [0,2] and the final 2 at [0,3], completing all constraints.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Sum-0 Forces Twin Zeros
The sum-0 region [2,0]/[2,1] can only be satisfied by 0+0. Since no domino supplies two 0s in these cells directly, two different zero-bearing dominoes are required. The available zeros are in [0,0], [2,0], [0,1], and [5,0].
2
Step 2: The Double-Zero Equals
The equals region [2,3]/[3,3] demands identical pips. The only domino with a matching pair is [0,0], so place it there with both cells 0. This removes one zero-bearing domino from the pool, clarifying the options for the sum-0 region.
3
Step 3: Solving the Sum-0 Pair
With [2,0] and [2,1] still needing zeros, place [2,0] domino with 2 in [3,0] and 0 in [2,0]. Then place [5,0] with 5 in [2,2] and 0 in [2,1]. The 5 in [2,2] satisfies the greater-3 constraint there.
4
Step 4: Chain Through Sum-5 and Equals
The sum-5 region [3,0]/[3,1] now contains a 2 in [3,0]. To reach 5, [3,1] must be 3. The [3,1] domino provides 3 at [3,1] and 1 at [3,2]. The equals region [3,2]/[4,2] forces [4,2] to be 1, so the [0,1] domino places 1 there and 0 at the empty cell [4,1].
5
Step 5: Top Row Resolution
The less-2 cell [0,0] needs a value <2, so it takes the 1 from [1,2] domino, placing 2 at [1,0]. The sum-5 [1,0]/[1,1] now has 2 + ? = 5, requiring 3 in [1,1]. The [3,4] domino gives 3 to [1,1] and 4 to [1,2], fulfilling the greater-3 on [1,2]. All dominoes are placed.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: The 2x2 Six-Block
The four-cell equals region at [3,2]/[3,3]/[4,2]/[4,3] requires identical pips. Only pip 6 appears in enough dominoes (six separate dominoes contain a 6) to cover four distinct cells, so this entire block is 6. All dominoes carrying a 6 must be strategically placed to cover these and other equals-6 regions.
2
Step 2: Double 2s and Double 5s
The three-cell equals region for 2s at [0,4]/[1,4]/[1,5] forces those cells to 2. The [5,2] domino can place 5 at [0,3] and 2 at [0,4], instantly fixing the 5-equals pair [0,2]/[0,3] to 5 and seeding the 2-equals. The [2,2] domino then covers [1,4]/[1,5] with two 2s.
3
Step 3: Sum-1 and Sum-0
The sum-1 cell [1,2] must be 1, and the sum-0 row [2,2]/[2,3]/[2,4] forces 0s. The [0,0] domino fills [2,3]/[2,4] with 0,0. The [0,6] domino fills [2,2]/[3,2] with 0 and 6 (the 6 joins the block). Then the [1,5] domino places 1 at [1,2] and 5 at [0,2], completing the 5-equals.
4
Step 4: Sum-10 and Greater-3
The sum-10 region [4,0]/[5,0] requires 4+6. The only available 6 not yet placed is in the [4,6] domino, so place it with 4 at [4,0] and 6 at [5,0]. The greater-3 cell [4,1] now needs a value >3, so the [5,6] domino supplies 5 at [4,1] and 6 at [4,2] (filling another block cell).
5
Step 5: The Triple-3 Equals
The equals region [5,2]/[5,3]/[6,2] demands three 3s. The [3,3] domino provides two 3s at [5,2] and [6,2]. The [6,3] domino (with 6 and 3) places 3 at [5,3] and 6 at [4,3], completing this cluster and the 6-block.
6
Step 6: Mopping Up 6s, 5s, and 1s
The remaining equals-6 region [6,3]/[7,2]/[7,3] is resolved by [6,6] at [6,3]/[7,3] and [2,6] at [7,2]/[8,2] (6 and 2 for the sum-2 at [8,2]). The equals-5 pair [7,4]/[8,4] takes [5,5] for two 5s. Finally, the equals-1 block [3,4]/[4,4]/[5,4]/[6,4] receives [1,1] at [5,4]/[6,4] and [0,1] at [4,5]/[4,4] (0 in sum-0, 1 in equals), satisfying all constraints.

💡 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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