Carnival - 3rd riddle: Find the odd one out

[Allison]

Dear Magoians,

On the last day we just want to make ourselves comfortable with a big mountain of doughnuts and let the event end comfortably.
Find the one 'Berliner' of each group of 4 (sporting the same 'stickers') that is different from the other three and form the solution word from the letters found.

Hint: click on the picture to get an enlarged view.


For the last task, please send the solution with the correct grouping of the 4 doughnuts, the number of the doughnut that differs from the others and, of course, the solution word to us by 22:00 server time (=20th Feb, 5:00 UTC) as a forum PN to dragonholt.

You can send it as a picture, but also in numerical form:
e.g. Group a) 1 - 2 - (3) - 4
Don't forget the solution word
Also, if you can’t find the word, don’t worry and send in the rest anyway because finding the correct groups and differing doughnuts also counts for some oft he prizes.

Please do not post the solution openly. Neither in the forum, nor in one of the public chats in all worlds.
In addition, please adhere to the known forum rules.

[Allison]

Now, for the last solution - I'm very impressed! All participants of this round got their answers correct and even found the solution word. And here I am, thinking that this one was a really tough one...

The groups with the odd ones out linked as follows:

1 - 8 - 14 - 16
2 - 5 - 6 - 10
3 - 7 - 13 - 15
4 - 9 - 11 - 12

The solution word came together as natural lights

For each correct group with the odd one out correctly identified, you could earn 150 Moon Mice - up to 600 in total.
For the correct solution word, you won 30... ah... well - what do you think would have fit? I guess you'll be able to guess.