NYT Pips Hints & Answers for August 13, 2026

Aug 13, 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!

Click here to play today's official NYT Pips game first.

Want hints instead? Scroll down for progressive clues that won't spoil the fun.

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🎲 Today's Puzzle Overview

Right from the start, today's NYT Pips easy puzzle, crafted by Ian Livengood, feels like a friendly handshake. The grid is compact and the domino set tiny—just five pairs. You’re immediately drawn to a small cluster of cells in the bottom row all forced to hold the same number, and another equals pair that must share a big value, leaving little room for doubt. The constraints are so direct that each placement cascades without much backtracking.

Moving to Rodolfo Kurchan’s medium grid, the pressure ticks up a notch. Now you’re juggling equals regions that lock neighboring cells together and sum regions that demand exact arithmetic. One sum-10 area that stretches across three cells will likely be your first head-scratcher: you know it needs three tiles, but only one specific domino can cover two of those cells while respecting the equals link nearby. After that, the domino choices narrow down elegantly, turning each constraint into a stepping stone.

Kurchan’s hard puzzle is where the NYT Pips formula truly shines. A sprawling grid hides a web of sum-15 and sum-5 regions that overlap and interconnect in devilish ways. A less-3 restriction at the center becomes an anchor, forcing low numbers that ripple outward into the larger sum regions. The equals rule on two vertical cells becomes a pivot, and you’ll find yourself carefully pairing dominos with 6s and 3s to satisfy sums that must coexist with nearby less-than limits. It’s a gratifying solve that demands both arithmetic precision and spatial reasoning.

💡 Progressive Hints

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

💡 Hint 1: Spot the commonality
Seek out a region where several cells share the same number—this is your key to unlocking the puzzle.
💡 Hint 2: Look to the bottom row
In row 1, the three rightmost cells sit in an equals region. That trio can only be filled with a pair of 1–1 dominos.
💡 Hint 3: Complete the chain
Place the [1,1] domino across [1,6] and [1,7], and the 1 from [5,1] at [1,5]; then send the 5 from that same domino to the empty cell above at [0,5]. The equals pair at [1,2]–[1,3] becomes two 6s from [6,0] and [3,6]; the 0 from [6,0] lands in the less-2 cell at [0,2], and [3,6]’s 3 goes to [1,4]. Finally, [0,0] covers [1,0]–[1,1] to satisfy the less-2 region there.
💡 Hint 1: Focus on a sum
Zero in on the region that demands two cells add up to exactly ten—it will dictate the biggest domino in the set.
💡 Hint 2: The corner sum-10
The sum-10 area at [1,4] and [2,4] needs a 6, which only the double-six can deliver, so that domino must land with a 6 in [2,4] and [3,4] above or below.
💡 Hint 3: Build outward
After placing [6,6] vertically over [2,4] and [3,4], set [1,4]=4. Fill it with the [4,4] domino covering [1,3]–[1,4]. Use [3,3] for [0,2]–[1,2] to complete the sum-10 across columns 2–3. Then [1,2]’s 2 and 1 satisfy [0,1] and [0,0]; [2,0]’s 2 and 0 handle [1,1] and [1,0]. [0,3]’s 0 at [2,0] matches the equals requirement, sending its 3 to [2,1]; finish with [5,4] giving 4 to [2,2] and 5 to [2,3].
💡 Hint 1: Sum-driven start
Search for a sum region that, because of its target number, can only be achieved with a specific combination of high-value doubles.
💡 Hint 2: Lower right corner
The sum-15 region in the lower right corner spanning [6,3], [7,2], and [7,3] demands a double-6 and a 3—the only way to reach 15 with available dominos is 6+6+3.
💡 Hint 3: Lock the doubles
Placing the [6,6] vertically over [6,3] and [7,3] locks those cells. That leaves [7,2]=3, which comes from the [1,3] domino; that domino’s 1 then must go to the less-3 cell [7,1].
💡 Hint 4: Equalize the center
Now shift focus to the equals region at [3,3] and [4,3]. With the [6,6] used, you still need two 6s there—one comes from the [6,0] domino (sending its 0 to the less-3 cell at [2,3]), the other from the [1,6] domino (with its 1 occupying the less-3 cell at [5,3]).
💡 Hint 5: Full unraveling
Complete bottom right: [6,6] at [6,3]/[7,3], [1,3] at [7,1]/[7,2] (3 at [7,2], 1 at [7,1]). Then equals: [6,0] at [3,3]/[2,3] (6/0), [1,6] at [5,3]/[4,3] (1/6). Sum-15 in first row: [4,6] at [0,1]/[0,2] (4,6), [5,2] at [0,3]/[1,3] (5,2), [4,2] at [0,4]/[0,5] (4,2)—that gives [0,2]=6, [0,3]=5, [0,4]=4, [0,5]=2, [1,3]=2. Sum-15 left column: [4,5] at [5,0]/[4,0] (4,5), [6,3] at [6,0]/[7,0] (6,3) gives [4,0]=5, [5,0]=4, [6,0]=6, [7,0]=3. Remaining: [0,3] at [7,5]/[6,5] (0,3), [2,2] at [7,6]/[7,7] (2,2), [5,3] at [6,8]/[7,8] (5,3), [2,3] at [4,8]/[5,8] (2,3), [0,0] at [1,8]/[2,8] (0,0).

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Aug 13, 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 13, 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 13, 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: Three equal in a row
The equals region spanning [1,5], [1,6], and [1,7] forces identical numbers across three cells. With only five dominos, the only way to get three equal pips is to use the double-1 [1,1] for two cells and borrow the 1 from the [5,1] domino for the third.
2
Step 2: Place the ones
Lay down the [1,1] domino horizontally covering [1,6] and [1,7], setting both to 1. Then place the [5,1] domino with its 1 at [1,5] and its 5 at the empty cell directly above, [0,5].
3
Step 3: Match the sixes
Next, the equals region [1,2] and [1,3] must hold equal values. The only high equal pair available comes from the 6 in [6,0] and the 6 in [3,6]. Place [6,0] so its 6 sits at [1,2] and its 0 at the less-2 cell [0,2] (0 is allowed). Then place [3,6] horizontally with its 6 at [1,3] and its 3 at the empty cell [1,4].
4
Step 4: Fill the less-2 pair
The less-2 region at [1,0] and [1,1] requires numbers below 2. The final domino [0,0] fits perfectly—place it there, setting both cells to 0. The puzzle is complete.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Corner sum-10
The sum-10 region at [1,4] and [2,4] must total 10. The only way to get a 6 into that pair is with the double-6 domino. Place [6,6] vertically over [2,4] and the empty [3,4], making [2,4]=6 and [3,4]=6. That forces [1,4] to be 4.
2
Step 2: Double-4 placement
To supply the 4 at [1,4], the double-4 domino must cover that cell. Place [4,4] horizontally across [1,3] and [1,4], so both become 4.
3
Step 3: Completing the first sum-10
Now the sum-10 region [0,2],[1,2],[1,3] already has 4 at [1,3]. The remaining two cells [0,2] and [1,2] must sum to 6. The double-3 domino fits perfectly—place [3,3] vertically at [0,2] and [1,2], giving 3+3.
4
Step 4: Equals and the left side
The equals region [0,1],[1,1] needs identical numbers. Use the [1,2] domino: its 2 at [0,1] and 1 at the empty [0,0]. Then place the [2,0] domino with its 2 at [1,1] and 0 at [1,0]. The equals region [1,0],[2,0] now forces [2,0]=0, so use the [0,3] domino with 0 at [2,0] and 3 at [2,1].
5
Step 5: Sum-7 finale
The sum-7 region [2,1],[2,2] now has 3 at [2,1], so [2,2] must be 4. The remaining domino [5,4] supplies that 4 at [2,2] and places its 5 at the empty [2,3]. All constraints satisfied.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Bottom right sum-15
The sum-15 region [6,3],[7,2],[7,3] can only be achieved with two 6s and a 3. Place the [6,6] double vertically at [6,3] and [7,3], yielding 6+6. This forces [7,2]=3, which must come from the [1,3] domino—place its 3 there and its 1 at the adjacent less-3 cell [7,1].
2
Step 2: Central equals
The equals region [3,3] and [4,3] demands two 6s. Use the [6,0] domino: put its 6 at [3,3] and its 0 at the less-3 cell [2,3] (satisfying the [1,3],[2,3] less-3 region). Then place the [1,6] domino, giving its 6 to [4,3] and its 1 to the less-3 cell [5,3].
3
Step 3: Top row sum-15
Now tackle the top row sum-15 region [0,2],[0,3],[0,4]. The [4,6] domino gives 4 at [0,1] (greater-3 requirement) and 6 at [0,2]. The [5,2] domino puts 5 at [0,3] and 2 at [1,3] (under the less-3 limit). Finally, [4,2] domino sets 4 at [0,4] and 2 at [0,5] (also less-3). Sum is 6+5+4=15.
4
Step 4: Left column sum-15
The left column sum-15 region [4,0],[5,0],[6,0] requires 5+4+6. Place the [4,5] domino with 5 at [4,0] and 4 at [5,0]. Then use the [6,3] domino: 6 at [6,0] and 3 at [7,0], which satisfies the sum-3 single cell at [7,0] exactly.
5
Step 5: Mop up the sums of five
For sum-5 region [6,5],[7,5],[7,6]: place [0,3] with 0 at [7,5], 3 at [6,5]; then [2,2] with 2 at [7,6], 2 at [7,7]. Sum-5 region [7,7],[7,8] uses [5,3] domino: 3 at [7,8] and 5 at [6,8] (single-cell sum-5). Place [2,3] at [4,8]/[5,8] for 2+3=5, and [0,0] at [1,8]/[2,8] as empties. All grids complete.

💡 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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