u/dje91090

P/p as in this puzzle is poop in my opinion. It somehow made the hard ones earlier in the week seem like a walk in the park. I just got absolutely stumped trying to come up with a strategy to attack this puzzle. Below is the mess of a walkthrough I came up with (bear in mind I tried to avoid pip counting at all costs, which may be the easier solution?). Would be very interested to know how others solved this (other than trial and error). Would also be curious to know how the former guide master would have attacked this (if he didn't hold a grudge against us all).

1. >!There are five doubles (0-0, 3-3, 4-4, 5-5 and 6-6). By forced placement, each of the three 3c= must contain one.!<
2. >!The only whole domino with a sum of 3 is the 1-2.!<
3. >!Very few tiles can fit in the bottom of the 3c9. The 6-6, 6-5, 6-4 and 5-5 are all too large, and the 0-0 and 0-1 are too small. The 6-3 and 5-4 would need to be followed by a 0-x (with x>4), which doesn't exist. The 4-4 would need to be followed by a 1-x (with x>4) which doesn't exist. The 2-5 would need to be followed by a 2-x (with x>4) which doesn't exist. This leaves three possibilities: 3-3 (finished with 3-6), 3-4 (finished with 2-5), or 1-2 (finished by either 6-6 or 6-5).!<
4. >!Note that in all of these combinations (3-3-3, 3-4-2 or 1-2-6), the 3c9 is composed of two numbers ≤3.!<
5. >!There are 11 total "small tiles;" three 0s, two 1s, two 2s and four 3s. The 2c<3 requires two. Each 2c3 require two. Two go into the 3c9. Three remain.!<
6. >!The only possible places that can take a number ≤3 are the 1c>0, discard, 3c= or [one more in the] 3c9.!<
7. >!Let's assume for a moment that the 3c= takes a low number. The only ones with enough are 0s or 3s.!<
8. >!If the 3c= is 0s, all 0s are booked, which means that both 2c3s are 1+2, and now all numbers ≤2 are booked and we can't complete the 2c<3.!<
9. >!If the 3c= is 3s, there's one 3 remaining. Which means that the 3c9 can only be 3-4-2 or 1-2-6. Either way, one 2 is booked. This means that one of the 2c3s has to be 0-3, which books the last 3, and forces the 3c9 to be 1-2-6 (with the 1-2 in the bottom). The other 2c3 would have to be 1+2 also. This books all 1s, all 2s and all 3s. So the 2c<3 has to be 0+0. And the 2c<3 can't be made of two tiles (because then the left half of either the 0-0/0-1 goes into the 2c>10, which doesn't work), so it would be the whole 0-0. BUT then this fails, because above this the 2c3 would need a whole domino, and we already used the 1-2.!<
10. >!So therefore a 3c= is not made from a number ≤3. So they are made of 4s, 5s and 6s.!<
11. >!This also means that the three remaining low numbers have to go into the 1c>0, discard and 3c9.!<
12. >!Which means that the 3c9 is made of three low tiles, and this has to be 3-3 with the 3-6 above it. Three 3s are booked, one remains.!<
13. >!The 2-5 is now more limited as to where it can go. It can go on the 1c>0-3c= border (on the left in the capital P), the discard-3c= border (on the top of the capital P), or the 2c<3-2c>10 border. No matter where, one 2 is therefore going into a place that is NOT a 2c3. So one 2c3 is a 0+3 (which books the last 3), and one is a 1+2 (which books the other 2). This leaves us with two 0s and one 1 to fill out the remaining unfilled spaces of the 1c>0/discard/2c<3.!<
14. >!If the 2-5 goes in either slot within the capital P, it would require the 2-1 to go into the 2c3. But then this would be followed with either a 0/1 going into a 3c=, and since there is none with a half of 4/5/6, this fails.!<
15. >!So the 2-5 goes on the 2c<3-2c>10 border in the lowercase p.!<
16. >!The 2c<3-2c3 border can be the 0-0 or 0-1. If it's the 0-1, the 2c3 is finished with the 2-1 going into the 3c=, which fails. So it's the 0-0, followed by the 3-4 and then the 4-4.!<
17. >!The 0-1 now only has one place to go, it's on the discard/2c3 border in the capital P with the 0 in the discard. This is followed by the 2-1 beneath it into the 1c>0.!<
18. >!There are two 4s remaining, and they must go into the two 1c>3 slots. The 4-6 is on the 1c>3-2c>10 border to finish the lowercase p, and the 4-6 is on the 1c>3-3c= in the capital P.!<
19. >!Finish this 3c= with the 6-6.!<
20. >!The 1c>4-3c= on the right of the capital P is therefore the 5-6, and finished with the 6-6.!<

Q.E.D. I say good riddance to this puzzle.

Nice one today IMO. Hope you all don't "take the L" and solve it yourselves!

Pre-placement:

1. >!The 3c<2 is theoretically either 0+0+0 or 0+0+1. We only have two 0s, so it's the latter. All 0s are booked, and one 1 remains.!<
2. >!The 6c≠ (the lowercase l) therefore contains one of each pip from 1 through 6. All 1s are booked, and one 2 remains.!<
3. >!The 1c<3 is therefore a 2. All 2s are now booked.!<
4. >!This means that the 2c7 has to be 3+4, and since there isn't a whole 3+4 domino, it's two tiles, with the bottom one going horizontally into the 3c<2.!<
5. >!We started with six 3s, four 4s, four 5s, six 6s, but have so far booked two 3s, two 4s, one 5 and one 6. This leaves us with four 3s, two 4s, three 5s and five 6s.!<
6. >!The only number left with enough to satisfy the 5c= is 6s. All 6s are booked.!<
7. >!With the 6s gone, the 2c10 is therefore 5+5, and since there isn't a 5-5 tile, it's made of two vertical tiles. One 5 remains.!<
8. >!We haven't yet accounted for the two 2c= or the three 1c>3 cells. With what we have remaining, clearly both 2c= are made of 3s, and the 1c>3s are 4, 4 and 5.!<

Placement:

1. >!Place the 5-3 on the 2c10-2c= border.!<
2. >!To its left, there is no 5-5, so it's the 5-4.!<
3. >!The 3-0 can only go on the 2c7-3c<2 border.!<
4. >!The 5-0 can only go on the 1c>3-3c<2 border.!<
5. >!Place the 1-6 to its left on the 3c<2-5c= border.!<
6. >!There's no 4-3, so the 2c7 is finished with the 4-6 horizontally into the 5c=.!<
7. >!The only place for the 3-3 to go is wholly within the 2c= above this.!<
8. >!The 6-6 goes to its right.!<
9. >!We have two dominos with 4 left to use, with the 4 slotted for the 1c>3. But the 1-4 doesn't work because the 1c<3 is a 2. So it's the 4-2 vertically. (If you tried a different arrangement here, you'll see that you can't finish the 6c≠).!<
10. >!Finish the capital L with the 6-3.!<
11. >!The 1-4, 2-3 and 5-6 all go in the 6c≠ (the lowercase l) in any order and any orientation (hence, as u/harlows_monkeys pointed out, there are 48 possible "different solutions" to this puzzle).!<

Happy 1 year anniversary to this game! We get an appropriately-themed puzzle today.

1. >!There are six 1s. These are all booked into the five 1c1s, and the 3c1.!<
2. >!Each 2c>11 is 6+6. The 2c11 is 6+5. All 6s are booked.!<
3. >!The 2c>11-1c>1 in the "r" must be the 6-2.!<
4. >!There is no 6-5 whole domino, so the 2c11 is made of two dominos. This forces tile placement: there is a domino on the 2c11-3c1 border, the 1-1 above it, and the 6-6 to the right wholly in the 2c>11, and then a domino on the discard-2c11 border.!<
5. >!The only place where the 1-6 can go, therefore, is on the 3c1-2c11 border.!<
6. >!Finish the 3c1 with the 0-0.!<
7. >!There are six 0s remaining. Where can they go? If they went in either of the 3c11s, we'd need to finish those cells with 6+5 (but all 6s are already booked). And we just used the 0-0 in the bottom left of the "A", so the 2c= on the bottom right of the "A" can't be 0s. So this means that both two 2c=s in the "E" are 0s, and both discards are 0s.!<
8. >!Finish the "r" with the 6-0.!<
9. >!The 2c11-discard is the 5-0.!<
10. >!The 1c1-2c= is the 1-0.!<
11. >!The other two 1c1s in the "Y" are the 1-4 and 1-5 going into the 3c11 (in either position).!<
12. >!This 3c11 is finished with the 2-2 into the 1c>1 (the 2-0 doesn't work).!<
13. >!The 5-5 is in the 2c= on the bottom right of the "A".!<
14. >!The 3-5 is the only tile that works in the middle of the "E", but which way does it go? If the 3 is in the 3c11 half and the 5 in the 1c>1 half, then we'd need two halves with a sum of 8 to finish the 3c11, which we don't have.!<
15. >!Place the 3-5 in the middle of the "E", with the 3 in the 1c>1, and the 5 in the 3c11.!<
16. >!The 3c11 is finished with the 4-0 and 2-0 (can be in either position).!<
17. >!The 2c=-1c>1 at the top of the "E" is therefore the 0-3.!<

Noticed there wasn't a guide yet for today and this is not a nice one by any means, but it's the best I got. Happy to hear other people's strategies in the comments.

Pre-placement:

1. >!2c>10s can either be 6+6 or 6+5. 2c>9 can be 6+6, 6+5, 6+4 or 5+5. There are three 6s, two 5s and three 4s available.!<
2. >!There are no dominos with both halves large enough to satisfy any of these as a single domino, so they're all made of two dominos. Which also means by forced placement that all the 2-celled regions are made of two dominos.!<
3. >!This also places a double in the bottom of the 3c=, which is either 1-1 or 2-2.!<
4. >!If either 2c>10 is 6+6, the other would need to be 6+5, and then we can't make the 2c>9. So both 2c>10s are 6+5, and the 2c>9 is 6+4. This books all the 6s, all the 5s and one 4.!<
5. >!One 4 is booked for the 1c4. One 4 remains.!<
6. >!There are five 0s. Two are booked for the 2c0. Three remain.!<
7. >!There are two 3s, five 2s and six 1s.!<
8. >!2c<3 can theoretically be 0+0, 0+1, 0+2 or 1+1.!<
9. >!The only places that the last remaining 4 or either 3 can go are in the 2c4s or in the 4c4.!<
10. >!Let's start with where the 4 can go. If the 4c4 was in fact 0+0+0+4, then all 0s and 4s are gone. So the 2c<3 has to both be 1s. This leaves four 1s. There aren't enough 3s for all four of them to go into the four 2c4s, so the 3c= is 1s, leaving one 1 to go with one of the 3s in a 2c4. But then we'd have a spare 3 with nowhere to place it, so this fails.!<
11. >!So at least one of the 2c4s is 4+0. All 4s are booked, and two 0s remain.!<
12. >!Now, where can the 3s go? If the 4c4 took one of the two 3s, it would be made of 3+1+0+0, which would book all the 0s, and leave us with five 1s and one 3. Once again, this would force both sides of the 2c<3 to be 1s, leaving three 1s, which would have to go into the 3c= and again leave behind a spare 3.!<
13. >!So therefore, both 3s go into a 2c4 somewhere. This books both 3s; four 1s remain.!<
14. >!To recap, there are four 2c4 spots; we already know three of them (0+4, 3+1 and 3+1). The last one must be 2+2. Three 2s remain.!<
15. >!We haven't yet determined what completes the 2c<3, 3c= and 4c4, and we have three 2s, four 1s and two 0s.!<
16. >!Where can these three 2s go? Let's assume for a moment they don't go into the 3c= (meaning that the 3c= is 1s, and would leave one 1 remaining).!<
17. >!The most the 4c4 can take is two 2s (along with the last two 0s). But with only a 2 and a 1 left over, the 2c<3 can't be completed.!<
18. >!If the 4c4 took only one two, 4c4 would be 2+1+1+0, but there are not enough 1s left to make this combination.!<
19. >!So, the 3c= is 2s, which books all the 2s.!<
20. >!This leaves us with four 1s and two 0s yet to be placed. Which means that the 4c4 has to be 1+1+1+1, and the 2c<3 are both 0s.!<

Placement:

1. >!Place the 2-2 on the bottom of the 3c=.!<
2. >!Above this can either be the 2-0 or 2-4 (the 2-6 is too large), which means that the this 2c4 is 0+4. The tile above into the 2c>9 has to be either a 4 or 6. There is no 4-4 or 4-6, so it has to be the 2-4 followed by either the 0-4 or 0-6 (both can work).!<
3. >!Whichever was not placed continues into the 2c<3 (either the 6-0 or 4-0).!<
4. >!With the 0-6 gone, the 2c<3-2c>10 is the 0-5.!<
5. >!The last 4 is on the 1c4-2c4 border, its the 4-1.!<
6. >!The 3-0 goes next to it.!<
7. >!The last 0, the 0-2, goes above it.!<
8. >!Finish the left side (the capital O) with the 2-6.!<
9. >!Let's move to the lowercase o. Place the 1-1 in the middle of the 4c4.!<
10. >!The 1-3 is the only domino small enough for the 4c4-2c4 border.!<
11. >!The 1-5 and 1-6 can be placed in either configuration to finish it up.!<

Toda¥'s hard pips is, at long last (after the last three brutal da¥s), quite simple IMO.

1. >!By forced placement, the 3c0 contains the 0-0 going horizontally in the middle of the "E".!<
2. >!Also by forced tile placement, the top of the blue 4c= in the "Y" is a vertical double, the top of the turquoise 4c= in the "N" is a vertical double, and the pink 2c= in the "E" is a horizontal double. There are only three available doubles; the 1-1, 4-4 and 6-6. All doubles are booked.!<
3. >!With the 4-4 booked, the only whole domino left that can fill the 2c8 is the 2-6.!<
4. >!3c>16 is either 6+6+6 or 6+6+5, and has to contain a whole domino vertically above the 0-0, followed by another domino going into the 1c>3.!<
5. >!The 6-6 is already booked. So the 6-5 is in there vertically, and finished into the 1c>3 with the 6-4.!<
6. >!There are now not enough 6s for either 4c=, so the 6-6 goes in the 2c=.!<
7. >!In the "N", there is a vertical tile on the border of the turquoise 4c=-1c>3. Of the available doubles (1s or 4s), there is no 1-x tile where x is >3. So the turquoise 4c= in the "N" is composed of 4s, and the blue 4c= in the "Y" is composed of 1s. !<
8. >!Place the 4-4 vertically in the top left corner of the "N", and the 4-5 vertically beneath it.!<
9. >!Place the 1-1 vertically on the top of the 4c= in the "Y".!<
10. >!The 1-2 is the only tile that can be on the 4c=-1c>0 border.!<
11. >!The bottom of the "Y" is then the 1-0.!<
12. >!The "E" is finished with the 0-2 (3c0-1c<3 border).!<
13. >!The only remaining tile for the 1c>4 in the "Y" is the 5-0.!<
14. >!The 2c0-1c>3 therefore has to be the 0-4.!<
15. >!Place the last remaining 0, the 0-3, above it, and the last remaining 4, the 4-3, to complete the "N".!<

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!

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!

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.!<