u/jxd73

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 need 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, and given the available pieces, 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, (2-5 and 2-6 are interchangeable) 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.

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.

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.

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.!<

I see some are having trouble with medium.

1. >!after working out the orientation of the pieces, we realize the red= are two pieces, the orange 8 and the purple = are self contained.!<
2. >!only two 0s, if one goes in the =/= then there'd be impossible to place the other 0, hence 0s are in the red=!<
3. >!Since the blue 3 has to "go up" (based on the work in step 1), it's the 0-3.!<
4. >!At this moment we have 5 unique digits, 2,3,4,5,6, but the 2 and 6 is a single piece, if used for the 8, then the size four =/= becomes invalid. Therefore the 8 is 4-4.!<
5. >!If we use the 5-5 in the purple =, then we'd be left with three 3s, there would be only two more tiles not in the =/= and one of which cannot be 3, making this configuration invalid. Thus 3-3 is in the purple=.!<

I tried working out the orientation but I do not think there is a single possibility.

>!Three 4s, all reserved. This rules out 0-4 for the purple 4 region.!<

>!eight 0s, seven reserved, all marked 0s have to link to an =, or a region less than 5.!<

>!The only double is 0-0, we can immediately place that in the orange 0 region.!<

>!1s, 2s, 3s, 5s, 6s need to make up four = regions, the largest is of size 4. Only 1s, 2s, 3s and 5s satisfy that condition.!<

>!6 cannot be in any >4 tiles because that would leave insufficient 6s to fill out another =, and at least one or two 6s with no where to go. (6 isn't eligible for the void because that requires attaching to a 4)!<

>!Since >4s are 5s, only three 5s left, the only digit of sufficient quantity for the red= are 1, 2 or 3!<

>!6 cannot be in the purple= because that area needs to connect to a 4!<

>!6 cannot be in the central blue= because that area needs to connect to a >4 which we established as 5!<

>!6 has to be in the vertical blue=!<

>!Because one tile from the region filled with 6 need to overlap with the red=, and there is no 3-6, we now know the red= is filled with 1s or 2s!<

>!Can it be 1? No, because no matter what, we need at least a least a 1 to make up the purple 3, the purple 4, and the teal 2.!<

>!Therefore, the red= contains 2s!<

>!Due to geometry, one piece needs to link the purple 4 area to the red =, with no 1-2, 2-2, and 2-4 taken elsewhere, it has to be 2-3, forcing 0-1.!<

>!Can 0-4 go in the top left, into purple 3 and red 4? If we do we'd need a 3 to complement the 0 and end up with two spare 3s and no where to put them. Therefore only place for the 0-4 is void and orange 4.!<

>!Now that we used up all the 0s, purple 3 is 1+2, teal 2 is 1+1. We are now out of 1s.!<

>!examine the red >4, it needs to be horizontal, otherwise there has to be a double to its left. Thus it has to be 5-3.!<

>!Purple = has to be 5s because it must connects with the last 4.!<

>!Red 4 at the top has to be horizontal too, there is no 1-4 so it's 2-4.!<

>!Recall that the vertical blue= are 6s, therefore 1-6 is below the 2-4.!<

From here the placement is largely forced.

This is quite straightforward.

1. >!0-0 is the only double and it must be in the blue equivalence area due to geometry.!<
2. >!with no more doubles the purple = area must be all horizontal.!< 2a. >!in which one tile must link to a 5!< 2b. >!the other must link to a >4, since there are no duplicate pieces, the >4 must be a 6.!< 2c. >!the only number that links with both 5 and 6 is 0.!<
3. >!there are three 2s but two are reserved so the orange = are 3s or 4s.!<
4. >!examine the 2-0, there is no place for it except the lower orange 0 and red 6.!<
5. >!where can the 2-1 go? Not on the purple 1 because the geometry doesn't work, it can only go on the blue 2 and teal 6.!<
6. >!2-3 can only be placed vertically.!<
7. >!can 5-4 be placed on the >3,4? No because then the 5-3 needs to be just below it, which means we have no 5s left to make the blue 6. 5-4 has to cover the 4,>4. !<
8. >!5-3 then has to go to the blue 6.!<
9. >! >3,3 is the 4-3!<
10. >!we need a 4 to make the red 6, so the orange = consists of 3-0 and 3-1.!<
11. the rest are forced.

chx_ hasn't posted so I'll give it a shot, it's much simpler than my guide makes it out to be.

Pre placement:

we have four 0s, three 1s, five 2s, seven 3s, five 4s, five 5s, three 6s. we have three doubles, 3-3, 5-5, 6-6.

11 has to be 5+6.

13 can be a) 2+5+6, b) 3+5+5, c) 3+4+6, d) 4+4+5, based on the available pieces

check the comment for screenshot.

>!0. we can right away deduce the orientation of some of the pieces, they are denoted by black lines (see pic)!<

>!1. the green 12 is obviously the 6-6.!<

>!2. the blue0 and >4 is has to be 0-5.!<

>!3. there are three 0s left, all had to be in the blue 0 region.!<

>!4. this means the teal 3 has to be 1 and 2, thus taking up all the 1s.!<

>!5. The teal 11 has to be 5+6 and we only have a single 6 left, the 6-2. !<

>!6. there are five 2s, three of which are already spoken for (see step 4), this means that the remaining 2s cannot be the orange 3=. Since there are no more 6s, 2+5+6 and 3+4+6 are now impossible, and 2s cannot be part of the green 13.!<

>!7. Since one domino has to overlap both the 0 region and the orange 3=, the 3= consists of either 3s or 4s.!<

>!8. given what we know now the green 13 is either 3+5+5 or 4+4+5.!<

>!9. In both the 3+5+5 and 4+4+5 scenario, another whole domino needs to fill the other two spots of the green 13 because we are constrained by the 0 and the 1 next to the 13.!<

>!9a). if 4+4+5, we'd have three 5s left (5-3, 5-5). This results in a contradiction, if 5-3 is used for the red 5 at the top, we'd have to use 5-5 in the teal 11, but as the 11 is surrounded by equivalents, we'd have no more 5s left to make it happen; OTOH if 5-5 is used at the top, we would run out 5s for the 11.!<

>!9b). ergo, the purple 2 and green 13 are 2-3 and 5-5. !<

>!10. 3-3 has to go to the blue equivalent on the bottom.!<

>!11. with all the doubles used, we can draw in the rest of the layout, denoted by red lines.!<

>!12. There is no 1-0, so the right size of the 3 is the 2-0 pointing down.!<

>!13. In the 11, can the 6-2 take the upper position pointing left? No, because we then need a 2-2, so 6-2 is lower pointing right, the 2 in the purple 2=.!<

>!14. this means the red 1 is the 1-2, pointing left.!<

>!15. only one 2 is left, 4-2, it has to point to the left.!<

>!16. 5-4 makes up the 11.!<

17. The rest are all forced, so I won't bother.

There was some strange words/phrases I have never seen before so I googled, and the AI summary revealed what the Connections category is.

I might have missed some lore items but is Thomyris supposed to be Darius' wife, with Vahram being her step-son? So if she wanted her own kid on the throne, why not just have a few children with Darius first, then murder Darius and Vahram? Surely the crown wouldn't pass to her but to another male relative.