NYT Pips Hints & Answers for October 13, 2026

Oct 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!

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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

The easy grid is a one-pass warm-up. Nearly every clue is a sum, and two of them are single squares that announce their own value the moment you look at them, so those tiles drop in without a single cross-check. The real pinch point sits along the bottom edge: a two-cell region with the smallest target on the board, which can only be built from the two tiles in the tray that carry a single pip. Solve that pair and the rest of this NYT Pips easy — the equals pair in the bottom-right corner, then the four sum-6 regions that fill the top half — unrolls in a straight line. Ian Livengood has left you a board that rewards scanning rather than staring.

The medium board, also by Livengood, asks you to read two kinds of clue at once and keeps you moving. Your first foothold is structural rather than numerical: the lone corner square in the bottom-right can only be reached from directly above, so you know that tile's shape before you know its value. The three-cell less-2 block in the bottom-left is the other anchor — three cells that all have to stay tiny, which is only possible if a double lands inside it. Once those two are in, the interlocking pairs do the work: a teal sum-3 pair that needs its second pip immediately, the purple less-4 pair directly to its left, and the green greater-8 pair that has to reach a big total from two modest pips. Work outwards from the corners and you will never need to backtrack.

Rodolfo Kurchan's hard grid is the day's real test, and it opens with an embarrassment of riches: seven separate sum regions, two equals regions and a three-cell unequal run that refuses to repeat a value. That bounty is also the trap — you have to keep track of which 6, which 4 and which 3 you have already spent, because the doubles get claimed early and nothing can be borrowed later. The bottom-right corner hands you the first freebie, a two-cell sum with only one possible tile, and from there the deductions march right to left: a sum-4 square that cannot afford the wrong 4 above it, two equals regions that between them swallow both doubles, and finally the triple-equals column on the far left, whose value is settled by a contradiction in the unequal run below it. Very little guessing, a lot of bookkeeping.

💡 Progressive Hints

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

💡 Three sums, one pinch point
Most of this grid is sums, and the tightest of them runs along the bottom edge with the smallest target on the board. Start there rather than with the equals pair in the bottom-right corner — a pair of cells that must add up to almost nothing narrows the tray far more than any other clue.
💡 Two 1s are all you get
The green sum-2 pair along the bottom edge is the anchor. Both of its cells have to hold a 1, and only two tiles in the tray carry one — the 6/1 and the 1/3. Those two tiles are committed to that pair, and the 3 and the 6 on their other halves already point at where the neighbouring regions go.
💡 Complete easy layout
The green sum-2 pair along the bottom takes both of the tray's 1s. The 1/3 tile stands upright in its left cell, 1 in the lower half and 3 above in the lower cell of the orange sum-6 pair; the 6/1 tile lies flat to its right, 1 as its left half in the green pair and 6 as its right half in the lower cell of the blue equals pair at the bottom-right corner. The orange pair then reads 3 above and 3 below, so the 3/6 tile lies flat with 3 as its right half in the upper cell of the orange pair and 6 as its left half in the teal sum-6 single cell. The equals pair takes 6 and 6: the 6/0 tile stands vertically with 6 in the upper cell of the blue pair and 0 in the lower cell of the pink sum-6 pair, and the 6/6 double lies flat across the top, its left 6 in the purple sum-6 single cell and its right 6 in the upper cell of the pink pair.
💡 Single cells first
Half of today's clues are single cells, and single cells are ruthless: a sum square announces its own value, and a greater-than square throws out every small pip in the tray. Take those before you touch the multi-cell blocks — and notice that the bottom-right corner square has only one neighbour at all.
💡 The corner that only looks up
The lone pink sum-2 square in the bottom-right corner can be entered from exactly one direction, so its tile must run vertically down the right-hand edge with its other half in the orange greater-3 pair directly above. That shape, plus a target of 2, strips the tray down to a very short list.
💡 Full medium layout
Right edge: the 0/2 tile stands vertically, 0 in the lower cell of the orange greater-3 pair and 2 in the pink sum-2 corner cell. The orange pair then needs a 4 above the 0, so the 4/2 tile lies flat across the top-right, its 2 as the left half in the upper cell of the teal sum-3 pair and its 4 as the right half in the upper cell of the orange pair. Teal needs a 1 under that 2: the 0/1 tile lies flat, 1 as its right half in the lower cell of the teal pair and 0 as its left half in the upper cell of the purple less-4 pair. That pair must stay under four, so the 4/3 tile lies flat with 3 as its right half in the lower cell of the purple less-4 pair (0 plus 3 is under four) and 4 as its left half in the lower cell of the green greater-8 pair. Green then wants 5 above the 4: the 1/5 tile lies flat with 1 as its left half in the top-right cell of the blue less-2 block and 5 as its right half in the upper cell of the green pair. The blue block takes the 0/0 double flat across its bottom two cells, giving 1, 0, 0 and a total of one. Finally the pink sum-3 square at the top: the 3/5 tile lies flat with 5 as its left half in the purple greater-3 square and 3 as its right half in the pink square.
💡 Two-cell sums have no slack
Two-cell sum regions are the fastest thing on this board: both cells are covered by a single tile, so the target tells you exactly what that tile has to add up to. Start with the most extreme of those targets and let their dominoes fall before you touch the equals regions or the unequal run.
💡 The generous corner
The green sum-10 pair in the bottom-right corner is the friendliest clue here. Exactly one domino in the tray reaches ten — there is no 5/5 to be had, so it has to be the 4/6 — and that tile stands upright, which means you know the shape of your first placement as well as its values.
💡 Follow the right edge down
The purple sum-4 square on the right edge shares a single tile with the teal greater-2 cell beneath it, and that lower cell has to hold something bigger than 2. The 4/6 is already spent, and both 2/4 and 4/1 would leave a pip of 2 or 1 down there — too small. That leaves the 4/3 tile: 4 in the upper half, 3 in the lower half.
💡 The two doubles split up
The top-right orange equals pair touches nothing but itself, so one upright tile fills it and equality forces a double — the 2/2 goes in, 2 above and 2 below. That leaves the 3/3 double for the blue equals column on the far left, where it covers the top two of the three cells. The column's third pip has to arrive later from a separate 3-tile.
💡 Complete hard layout
Top-left corner: the 4/1 tile lies flat, 4 as its left half in the purple sum-4 square, 1 as its right half in the pink less-2 square; next to it the 6/1 tile lies flat with 6 in the uncoloured free square and 1 in the teal sum-1 cell. Far-left column: the 3/3 double stands upright over the top two cells of the blue equals column, 3 upper and 3 lower; the 3/5 tile lies flat at the bottom of that column, 3 as its left half in the lowest blue cell and 5 as its right half in the left cell of the pink unequal run; the 0/6 tile lies flat beside it, 0 in the middle cell of the run and 6 in the right cell, so the run reads 5, 0, 6 — three different values. Bottom-left: the 2/4 tile stands upright, 2 in the orange less-3 square above and 4 in the purple less-5 square below. Top-right: the 2/2 double stands upright in the orange equals pair, 2 above and 2 below. Right edge: the 4/3 tile stands upright, 4 in the purple sum-4 square and 3 in the teal greater-2 cell beneath it. Middle: the 2/5 tile stands upright in the green sum-7 pair, 2 above and 5 below; the 2/6 tile stands upright in the blue sum-8 pair, 6 above and 2 below; the 4/6 tile stands upright in the green sum-10 pair, 4 above and 6 below. Bottom-right: the 3/1 tile lies flat in the pink sum-4 pair, 3 on the left and 1 on the right.

🎨 Pips Solver

Oct 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 October 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 October 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: Find the smallest target
The green sum-2 pair along the bottom edge is the tightest region on the board: two cells, and only 1 plus 1 gets there. The tray holds exactly two tiles carrying a 1 — the 6/1 and the 1/3 — so both of them are committed to that pair before you look anywhere else.
2
Step 2: Both 1s land
The 1/3 tile stands upright in the left cell of the green pair, its 1 in the lower half and its 3 above, in the lower cell of the orange sum-6 pair. The 6/1 tile then lies flat, 1 as its left half in the right cell of the green pair and 6 as its right half in the lower cell of the blue equals pair at the bottom-right corner.
3
Step 3: The orange pair and the teal square
The orange sum-6 pair already carries a 3 in its lower cell, so its upper cell must be a 3 as well. The 3/6 tile is the only tile left holding a 3, and it lies flat with 3 as its right half in the upper cell of the orange pair and 6 as its left half in the teal sum-6 single cell. One move closes the orange pair at six and the teal square at six.
4
Step 4: The equals pair and the last double
The blue equals pair needs two identical pips, and its lower cell already reads 6 from the 6/1 tile, so the upper cell must be 6 too. The 6/0 tile stands vertically, 6 as its lower half in the upper cell of the blue pair and 0 as its upper half in the lower cell of the pink sum-6 pair. Pink now holds 0 below, so it needs a 6 above — and the 6/6 double is the only tile still unplaced. It lies flat across the top, left 6 in the purple sum-6 single cell and right 6 in the upper cell of the pink pair.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: The corner that only looks up
The pink sum-2 square in the bottom-right corner has exactly one neighbour, the orange greater-3 cell directly above it, so its tile must run vertically down the right edge. Stack the 0/2 tile there: the 2 drops into the pink corner and the 0 stays above, in the lower cell of the orange pair, which now needs a big pip to push its total past three.
2
Step 2: The top-right pair locks
With a 0 sitting in the orange pair, the upper cell has to be a 4 for the region to exceed three. The 4/2 tile supplies it and lies flat across the top-right: 2 as its left half in the upper cell of the teal sum-3 pair, 4 as its right half in the upper cell of the orange pair.
3
Step 3: Teal needs a 1
The teal sum-3 pair now holds a 2 on top, so the cell beneath it must be a 1. The 0/1 tile lies flat, 1 as its right half in the lower cell of the teal pair and 0 as its left half in the upper cell of the purple less-4 pair.
4
Step 4: Keeping the purple pair small
The purple less-4 pair already contains a 0, so its lower cell has to stay tiny. The 4/3 tile lies flat with 3 as its right half in the lower cell of the purple pair — 0 plus 3 stays under four — and 4 as its left half in the lower cell of the green greater-8 pair.
5
Step 5: Green's big total, and the rest
The green greater-8 pair needs 9, and its lower cell is a 4, so the upper cell takes a 5: the 1/5 tile lies flat with 1 as its left half in the top-right cell of the blue less-2 block and 5 as its right half in the upper cell of the green pair. The blue block then needs its remaining two cells to be tiny, so the 0/0 double lies flat across them for a total of one. Finally the pink sum-3 square at the top takes the last tile — the 3/5 lies flat, 5 as its left half in the purple greater-3 square and 3 as its right half in the pink square.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: The bottom-right corner gives itself away
The green sum-10 pair in the bottom-right corner can only be filled by one upright domino, and no pair in the tray reaches ten except 4 plus 6 — there is no 5/5 to be had. Place the 4/6 tile vertically with 4 in the upper half and 6 in the lower half.
2
Step 2: The right edge falls next
The purple sum-4 square on the right edge and the teal greater-2 cell directly beneath it share one vertical tile. The 4/6 is spent, 2/4 would leave a 2 in the greater-2 cell and 4/1 would leave a 1 — neither is bigger than 2. So the 4/3 tile goes in: 4 above in the purple square, 3 below in the teal cell.
3
Step 3: The top-left corner and the free square
The pink less-2 square can only hold a 0 or a 1, and it has to share a tile with the purple sum-4 square beside it, which needs a 4. Of the tiles left, only the 4/1 carries a 4 with a small enough partner: 4 as its left half in the purple square, 1 as its right half in the pink square. The teal sum-1 cell has just one neighbour, the uncoloured free square, so the remaining 1 — from the 6/1 tile — lands there: 6 in the free square on the left, 1 in the teal cell on the right.
4
Step 4: The bottom row
The pink sum-4 pair in the bottom-right must be one horizontal tile, since its left cell's only other neighbour is already occupied by the 4/6. It needs 3/1 or 2/2 to total four, and the 2/2 double is needed by an equals region, so the 3/1 lies flat: 3 on the left, 1 on the right. Further left, the orange less-3 square and the purple less-5 square below it take the 2/4 tile — 2 in the upper orange cell, 4 in the lower purple cell.
5
Step 5: Green sum-7 takes the only fitting tile
Among the tiles still in the tray, only the 2/5 adds up to seven, so the green sum-7 pair in the middle of the board must take it. Place it upright: 2 in the upper half, 5 in the lower half.
6
Step 6: Two doubles decide the rest
The orange equals pair and the blue equals column both need repeated pips. Give the 2/2 double to the orange pair, 2 above and 2 below; the column's top two cells take the 3/3 double, 3 above and 3 below. The column cannot be 2s instead — the only leftover tile that could extend it would be the 2/6, and its 6 would clash with the 6 the 0/6 places in the unequal run, which forbids repeats. So the 3/5 lies flat at the bottom of the column, 3 as its left half in the lowest blue cell and 5 as its right half in the left cell of the unequal run, and the 0/6 completes the run: 0 in the middle cell, 6 in the right cell, giving 5, 0, 6 — all different. The blue sum-8 pair finishes it with the 2/6 standing upright, 6 above and 2 below.

💡 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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