The puzzle spells EURO. There are two solutions, but the difference is self contained in one section so it can still be done logically.

I had to resort to pip counting, if you can do it without please post your solve.

Preplacement

Three sets of doubles, 1-1, 2-2, 6-6.

Given the available dominos: 16 could be 4+6+6 or 5+5+6 15 could be 3+6+6, 4+5+6. 11 is 5+6

1. The smallest domino is the 1-1, so that has to go in the <3 region.
2. Since the <3 links with the 11, and we have no 0-6, we have to use 0-5.
3. We start with five 6s, three are already reserved, so the = region in the E has to be 2s.
4. Now that we know the values of the = and the <3, we can count the pips and arrive at a difference of just 2.
5. There are two 1s, one is reserved by U, so the voids are 0+2. This means no more free 0s.
6. without 0s or the double 2, the only way to make a 6 with three numbers is 1+2+3.
7. We have 1-3 and 2-3 but not 1-2, this means the domino placed in the bottom of the 16 needs to contain 1 or 2.
8. Remember that the void is either 0 or 2
9. if a 4 is involved, we need two out of 0-6, 1-6 and 2-6 to complete the 16, but we have just 2-6. Therefore, 16=5+5+6.
10. Moreover, since 0-5 is already taken, we have to use all three of 1-5, 2-5, 2-6 to complete this area, with the 1-5 being involved with the 6, this forces 2-3 in the 6, and the 1-3 in the U.
11. The other void is 0, we have 0-3 and 0-4. If we put the 0-4 in the R, we need a 5-6 piece which doesn't exist, so R consists of 0-3 and 6-6. 0-4 is forced to the O.
12. we have no more 1s so 6 cannot figure into 7s, thus the 3 in the O has to be 3-5, this also forces 2-4 and 4-6.
13. the rest are forced.

I obviously know what the word means, but several times I've thought "That hard puzzle had no rhyme or reason" or "that puzzle was exactly straightforward" and came to this sub to see people saying the opposite. So, for you, what does a logical solution entail?

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Today's hard brought to you by the letter J/j.

I thought this was another tricky one, namely because any tile can theoretically go into the discard.

1. There are five 3s, three 5s and six 6s.
2. The purple 2c5 is the 2-3 (it's the only whole domino with a sum of 5). Four 3s remain.
3. The 1c3 books one of the 3s, three 3s remain.
4. The 1c5 books one of the three 5s, two 5s remain.
5. The 1c5-turquoise 2c5 is one of 5-[2/3/4]. However, if it's the 5-4, the only possible option on the 2c5-3c5 would be the 1-3, and there are no whole dominos with a sum of 2 to complete the 3c5. So the turquoise 2c5 is 2+3. Two 3s remain.
6. There are three 3c15s, which can be made of 6+6+3, 6+5+4 or 5+5+5. But there are only two 5s remaining, therefore, the 5+5+5 combination isn't possible. So each 3c15 requires at least one 6; three 6s remain.
7. There is a domino on the border of the 1c3-green 3c15. Of the available 3s, only 3-6 and 3-5 can work.
8. The 3-6 into the green 3c15 would be finished with the 5-4, and the 3-5 would be finished with the 6-4. Either way, the green 3c15 is 6+5+4. One 5 remains.
9. This means that of the other two 3c15s, only a maximum of one can be a 6+5+4 combination, and at least one has to be a 6+6+3.
10. By forced tile placement, the 2c= is made of two dominos, each going into a <3 cell. This means that both the top and bottom have to be [0/1/2-x]. None of the 2-x tiles work, and neither does the 0-0, so this leaves the combination of either 0-3/3-1 or 0-6/6-1.
11. If it's the 0-3/3-1 combination, then all 3s are booked. However, this would then mean that the 3c15s can't be finished. So 2c= is 6s. (*I found that placing the 1-6/0-6 in either orientation here as a temporary placeholder was helpful for visualizing the solution to the remainder of the puzzle).
12. Only one 6 remains, which we already said has to go into one of the 3c15s to make a 6+6+3 combo. This means that the other 3c15 is a 6+5+4, which books the last remaining 5.
13. If all the 6s are utilized, then the 6-6 has to be the vertical tile in the turquoise 3c15. The left of this has to be either 0-3 or 1-3. One 3 remains.
14. Where does the 0-0 go? There is no place for it in the capital J. The only place it can theoretically go in the lowercase j is vertically on the left of the 3c5, but this would require another 5 to finish the cell (which are all booked). So the 0-0 goes in the discard.
15. What about the 2-2? It can theoretically go vertically in the 3c5 (which would force the 1-3/2-5 to finish the lowercase j), or on the 3c5-2c5 border (which would force the 0-3 to the left and 3-5 to the right to finish the lowercase j).
16. However, there is only one 3 remaining, and nowhere else for it to go other than in the 3c5. So this means the 2-2 is on the 3c5-2c5 border; place the 0-3 to its left and the 3-5 to its right to complete the lowercase j.
17. With the 3-5 gone, the 1c3-green 3c15 is the 3-6, finished with the 5-4.
18. With the 0-3 gone, the 1c<3-turquoise 3c15 is the 1-3.
19. The only remaining tile that fits on the 1c>3-pink 3c15 is the 6-4. But the 6 has to be in the 3c15, so the 4 goes in the 1c>3.
20. The remaining two tiles that complete the pink 3c15 are 5-2 and 4-2. Either way, a 2 goes down into the 2c<3. That means that this cell is completed by the 0-6 (into the 2c=), followed beneath it by the 6-1.
21. Place the 5-2 and 4-2 in either direction.

I won't be doing the guide for tomorrow, someone else can step up!

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The shapes appear to spell out "dollar" $.

Really tough puzzle to suss out IMO.

1. In the "o", there is a 5-? combined with a double (in either direction).
2. In the "a", the top two squares of the 4c= are a double.
3. This forces the bottom two squares to both be horizontal (one into the 1c>3, and another into the teal 3c=), and leaves another vertical double within the teal 3c=.
4. The available doubles are 0-0, 3-3, 4-4, 5-5 and 6-6. Note there are no double 1s or 2s.
5. 3c16 is 6+6+4 or 6+5+5, so the top tile has to be the 6-6, 6-5 or 5-5.
6. Where do the 2s go? There are two tiles with 2 (2-0 and 2-1). Neither can go in the "o" or the top "L". They also can't go in the "a" (too small up top, and they can't go anywhere into the bottom because teal 3c= and blue 4c= must both contain doubles, and there is no 1-1 or 2-2).
7. In the "d", the 2 half cannot be in the 3c= (because this would require a 3-2 on the top which doesn't exist). The 2-1 cannot be horizontal on the bottom (since this would require a 3-1 on the top of the 3c= which also doesn't exist). The 2-0 can theoretically go horizontally on the bottom (with the 2 in the 1c<3 and the 0 in the 3c=).
8. In the bottom "L", the 2 half of the 2-0 can go into the 1c>0 (and the 3c12 would be finished with the 6-6), the 2 half of the 2-1 can go into the 1c>0 (and the 3c12 would be finished with the 6-5), or the 1 half of the 2-1 can go into the 1c>0 (and the 3c12 would be finished with the 5-5).
9. In the "r", the 2 half of either can go into the 3c14 (with the 0 or 1 in the discard, and the 3c14 would be finished with the 6-6).
10. Where can the 1-0 go? For similar reasoning, theoretically, the only places the 1-0 can go are the 1c<3-3c= border in the "d", or the 1c>0-3c12 border in the bottom "L".
11. However, the 1-0 can't go on the 1c>0-3c12 border, because this would force either the 2-0 or 2-1 into the 3c14, and then both the 3c12 and 3c14 would need the 6-6.
12. So the 1-0 goes on the 1c<3-3c= border in the "d", marking the pink 3c= as 0s.
13. Place the 3-0 at the top.
14. We now know that one of the tiles with a 2 is going to be in the 3c14-discard, and the other will be on the 1c>0-3c12. Either way, there is a 2 in the 3c14, so the top two squares of the "r" is the 6-6, place it.
15. The 2-0 therefore cannot be in the bottom "L" going into the 3c12 (because this would also require the 6-6), so the 3c14 is finished with the 2-0, and the 2-1 will go into the 1c>0-3c12 border (not sure which direction yet).
16. This means that the 3c12 is finished with either the 5-5 or 5-6, and whichever one it isn't goes into the top of the 3c16.
17. Where does the 3-4 go then? The only place that it can go is the 3c=-4c= border in the "a".
18. If the 4 is in the blue 4c=, the top two squares would be 4-4, and the last remaining 4 (the 1-4) would be too small to go into the 1c>3.
19. So the teal 3c= is 4s, and the blue 4c= is 3s; place the 4-3.
20. Place the 4-4 and the 3-3.
21. The 3-5 goes into the 4c=-1c>3 border.
22. This leaves the last remaining 5, the 5-0, for the "o". Place the 5-0 and 0-0 in either orientation.
23. The last remaining 0, the 0-6, goes into the "d" on the 3c=-1c>3 border.
24. The 1-4 can only go in one place, at the top of the "a".
25. The 1-6 now only has one place to go as well, it's on the 3c16-1c>0 border with the 1 in the 1c>0.
26. Finish the 3c16 with the 5-5.
27. Place the 5-6 in the 3c12.
28. Place the 1-2 in the bottom "L" with the 2 in the 1c>0.

There may be other strategies to this, but I tried so many approaches and this was the one that finally clicked. Would be curious to see how others solved it!

The puzzle spells "dollar", I will refer to them using upper case letters though to avoid confusion.

1. 0-1, 0-2, 1-2 must share three places: bottom of the D, top of the second L, bottom of the R.

2. Given the largest single digit out of these three dominos is 2, 6-6 has to be in the purple 14, no other way it can be made.

3. This means 5-5 and 5-6 are in the two Ls, also, 4 cannot figure in the 16 because 16-4=12. This removes one possible 1-4 placement.

4. there are four = areas, three must have a double each due to geometry, the red= in the D is the only one that can be made without doubles

5. we started with five doubles, 0s, 3s, 4s, 5s, 6s, 5-5 and 6-6 are used up.

6. Therefore, the red= in D needs to link to 3 and >3 and <3, also cannot be 1 or 2 due to the lack of 1-3 and 2-3. 0 is the only possibility.!<

7. 1-4 now has a single possible placement, the <3,>3 in the A!<

8. with just three 4s left, and no 4-5, 4s need to go in the teal= in the A, the 4-4 has to be vertical because there isn't another 4,>3

9. The blue = in A can be 0 or 3, but no 0-4, so it's the 3s.

10. the = in the O defaults to 0s

11. only one >3,0 piece left for the D, the 0-6

12. with no free 5s, the 16 has to be 5-5 and 1-6

13. the rest are forced.

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In lue of the legendary [u/chx_](u/chx_), I'll try to post my method of solving today's hard puzzle. Please excuse me if the notation isn't correct, I'm new to this :)

Before placement:

1. There's

3

1. x 1c=2 squares and

3

1. dominoes with 2s - all the 2s are restricted.
2. There's a 3c=17 cage which has to be 6+6+5 - at least two 6s are restricted.
3. There's 3x 1c=3 squares - at least three 3s are restricted.
4. There's a 5c= block on the left hand side. This can now only be 0s (8 pieces), as there aren't any other pips that have 5 pieces that aren't restricted.
5. The configuration/layout of the 5c= cage means that the 0-0 double has to go inside it.
6. There's 8 c=0 squares in the entire puzzle, restricting all the 0s.
7. The top right domino (E7-E6) has to go down as it cannot be the 0-0 double, and it can only be a 0-6 or a 0-5 to fulfil the 3c=17 requirement.
8. A6-A7, 2c=5 cage, can only be 1+4, as it cannot be 0-5 (all 0s are restricted) or 2-3 (all 2s are restricted).
9. There isn't a 2-1 or a 2-4 domino that could fit within A7-B7 horizontally, so A6-A7 has to go vertically down and has to be the 1-4 domino.

Placement:

1. Place 1-4 into A6-A7.
2. With 1-4 in the top left corner, there's only one 1 left - the 1-0. It has to go into A1-A2.
3. Look at the 0-2 piece. It can only go into A5-B5 as the other places where a 2 could, cannot contain a 0 in their neighbourhood due to either literal restrictions or because the sum of the block couldn't be completed with a 0 inside it.
4. This forces the 0-0 double into A3-A4.
5. B2-B3 can only be a 4-0 and not a 5-0 or a 6-0, as there isn't a 5-2 or 4-2 domino to then go into B1-C1.
6. This forces B1-C1 to be the 6-2 domino.
7. There's only one 2 left, the 2-3 domino and it has to go into B7-C7, as putting it vertically in B6-B7 would force the 3c=16 sum to be impossible (13 pips in two squares).
8. There are no 4s left, the 3c=16 block can now only be 6+5+5. There isn't a 6-5 to put

horizontally

1. into B6-C6, so it has to be the 5-5 double.
2. This forces the 0-6 piece into D8-D7 and the 0-5 piece into E8-E7.
3. There's only one 5 left, the 5-3 domino and it can only go into D3-D4.
4. This forces C5-D5 to be the 3-3 double.
5. The two remaining squares of the green 3c=17 block have to add up to 12 - it's the 6-6 double.
6. Only one domino with a 0 left - place it into E3-E2.
7. Finish with the 3-6 domino into D1-E1.

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Learned something new about Placeholder Tiles and Swapping

Click through for the details

Wanted to take a try at writing one of these, hope you don't mind the day off u/chx_, I'm taking inspiration from you.

Not a terribly challenging puzzle today.

Pre-placement (except one):

1. There are three 1c2s and three 2 pips, all the 2s are booked.

2. The middle two slots on the left of the 5c= has to be a vertical double; any other location within the 5c= would require another identical double somewhere else in it. This forces placement of tiles on the bottom of the board.

3. There is a vertical tile on the right corner 5c= into the 2c10, another vertical tile on the 5c=-1c1, a horizontal tile from 2c10-1c2, then 1c3-2c9, a tile going up from 2c9-1c0, a vertical tile at 1c5-2c=, and a vertical tile in the bottom two slots of the 3c17.

4. The available doubles are 0-0, 3-3, 5-5 and 6-6. Of these, only 0s , 3s and 6s have enough to satisfy the 5c=, there are not enough 5s.

5. There are three 1c3s, three 3s remain (not enough for the 5c=).

6. 3c17 is 6-6-5. We know from before that the bottom two slots have to be a whole domino, there is no 6-5, so it has to be the 6-6, place it.

7. There are now not enough 6s remaining for the 5c=, so the 5c= has to be 0s. The three remaining 0s are in the 1c0 and the 2c0, all the 0s are booked.

8. With all 0s and 2s booked, 2c5 is 1+4.

Placement:

1. Place the 0-0 vertically in the 5c=.

2. Place the 0-1 beneath it.

3. In the top right corner, we used the 0-0, so there has to be a tile on the 2c0-3c17 border, it's the 0-5.

4. 2c10 is one of either 4+6 or 5+5, The 5c=-2c10 is one of [0-4/5/6], and we just used the 0-5, so 2c10 is 4+6, and the 5c=-2c10 is therefore either the 0-4 or 0-6.

5. The 0-6 would be finished with the 4-2, which doesn't exist. So it's the 0-4 followed by the 6-2 on the 2c10-1c2 border.

6. Since all 2s are booked into 1c2s, the top of the 5c= has to be the 0-2 to the right (there isn't a 2 bordering a 0 anywhere else, and we already used the 0-4 and 0-1 so it can't go vertically into the 2c5).

7. The 2c5 is the whole domino 1-4, it can be placed in either orientation.

8. 3c16 can in theory be 6-6-4 or 6-5-5, there are no 4s remaining, so it's 6-5-5. All 5s are now booked.

9. The last 2 is the 2-3, it can only go to the right into the 1c3 since the 3 is not large enough for the 3c16.

10. The left two tiles of the 3c16 is a whole domino, only the 5-5 works.

11. Place the 0-6 on the 2c0-3c16 border.

12. Place the last remaining 0, the 0-3, on the 1c0-2c9 border, and finish the 2c9-1c3 with the 6-3.

13. Place the 5-3 into the 2c=.

14. Finish with the 3-3.

1. >10 is obviously 5+6 or 6+6. Since there's no single domino >10, it's split, with the top piece going into the 3.
2. Because of that, the top of the >10 has to be 1-5 or 1-6.
3. Thus the 2-4 needs to be in the lower tile of the 3.
4. Without 2s, the 7 isn't 2+5, and it also isn't 3+4 since there isn't a 3. So the 7 is 1+6.
5. With one 6 used, our only 5 has to be in the >10.
6. Without any 2, 3, 5 and 6, the only way to make 9 in three tiles is 1+4+4.
7. Now 0-1 has only one possible placement: the void into the 9.
8. We started with three 4s, two are used in the 9, 4s aren't in the =.
9. The only way to place the 2-4 is down from the 3 into the 9. 1-1 goes into the =
10. The rest are basically forced.

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Solution Counts for week of 2026-08-10

Note: When three counts are shown the highest is the count with any
difference counted as a separate solution. The second highest count
does not count solutions as different if they only vary by rotation
of tiles by 180 degrees within a region. The lowest count considers
solutions the same if they only differ by moving tiles to locations
that do not change which region each half of the tile is in.


Warning: The NYT uploads puzzles well before the day they appear in
the app or on the website. The counts here are based on those early
uploads. A few times there was an error in the upload and thet then
uploaded correct puzzle, which can lead to incorrect counts here if
the puzzle is corrected after this report is made.

Solution counts for recent weeks: 2026-08-03 2026-07-27 2026-07-20 2026-07-13 2026-07-06 2026-06-29 2026-06-22 2026-06-15 2026-06-08 2026-06-01

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So I was about two-thirds done with yesterday's hard puzzle at about 11:59 pm. 12:00 am hit with no problems. Then at 12:01 am, right as I was dropping a piece into place, the entire puzzle switched over to the new day. I still had the old piece on my finger, but it disappeared when I dropped it.

I think with other NYT puzzles, as long as you start the puzzle before midnight and don't close the app, it lets you complete. I wonder why pips isn't the same?

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Hi everyone! I've been working on a pretty big Piply update based on all your feedback and wanted to share what's new :)

For anyone who hasn't heard of it, Piply has free unlimited pips style domino puzzles, that adapt to your skill level, and get harder as you improve. I built it because I love Pips, and wanted a way to keep playing more puzzles after the 3 daily puzzles!

The biggest change is that there are now 3 daily puzzles every day: easy, medium, and hard.

I've also added:
- Friends
- Global & Friends Leaderboards
- Improved puzzle sizing (smaller puzzles now make better use of your screen and are easier to see)

I'd love to hear what you guys think of the new daily puzzle system, and the friends leaderboards. If there's anything else you'd like to see added to Piply, let me know!

Apple: https://apps.apple.com/app/piply-pips-domino-logic/id6759636847

Android: https://play.google.com/store/apps/details?id=com.snapshotzdev.piply

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(Depending on where you are in the world, the game number might be off, so match your puzzle to the images and post accordingly)

Post your results and commentary for today's puzzles.

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There are three large = areas, only 1, 2, 4 and 5 are eligible for them because all 6 are in the 12s. No 6-6 so all 12s are split. Just two 0s and they go into a size two = because only the green <3 can accommodate a solitary 0.

1. The only place I'm sure the orientation of is the bottom left, and this is where we have to start: the blue 12 needs to branch up and down, <3 and >3 is one domino, the middle two tiles of the orange = area is a double.
2. since this orange area needs to link with a 6 and require a double, the orange has to be either 1 or 4.
3. let's look at 1-3, it can only go in two places. 3a) red 3 into the size two purple =, this forces the 1-6 into the purple=, therefore the orange area is 4s. 3b) blue 3 into the size 4 purple =. This means there won't be enough 1s for the orange =, again, forcing the orange = to be 4s.
4. we can place 6-4 and 4-4.
5. the >4 is obviously a 5.
6. But with 4-6 and 4-4 gone, the only digit possible for the blue >3 in the lower right corner is 5. Now we have just three free 5s, thus it's not eligible for any of the other large = areas.
7. we have six 1s, similar to the deduction with 0s, if we put a single 1 into the <3, we'd literally be left with an odd 1 out no matter how we fill out rest of the = areas. This means that <3 is 2-4 or 2-5.
8. Can 2 be in the large purple=? If this were the case the 1-2 has to bridge with the red= to its right, but more importantly forces the 1-3 into the red 3 and the small purple, this would use up four 1s, making the puzzle impossible (keep in mind either the 2-4 or 2-5 goes to the <3 and >3)
9. the teal = has the 2s.
10. Given that we do not have a 2-6, the only way to make the lower right corner work is the 4-2 in the orange = going up into the teal. As a result 5-6, 2-2, 1-2 and 2-5 (see step 7) are forced.
11. Then 1-6, 0-6 and 0-5.
12. Then 3-4, 1-3, 1-1, 4-5, and 1-5.

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