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Off topic & small talk: another small lottery from gamble1 and me ;-) (Page 5)

Topic created on 14th Apr. 2020 | Page: 5 of 10 | Answers: 90 | Views: 16,716
bruffl
Expert
Iseedeadpeople wrote on 04/18/2020 at 2:36 pm

of course, only the beginning is shifted. at a, the count is then continued continuously

ok, thank you!

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bruffl
Expert
Task 1: Solution 5

Task 2: 97 iterated: 7

both added: 12 / iterated 3

it took a little longer than 5 minutes...🤔

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palme7777
Top Member
My solution now is

Task1: 4
Task2 :7

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Anonym
Because I can't edit my solution, here's an addition .

Task 1 , one-digit cross sum is 1
Task 2, result is 245

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mowolum
Elite
Task 1 : solution 6
Task 2 : 54 : iterated 9
added 15 : iterated 6

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Anonym
Did the math...

I calculated the Roman numbers M, L, C, then the remaining letters. There came out a 1

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schwede666
Top Member
How cool ! Have no idea and probably not even understood the task so right but when I see the response (times regardless of the fact that almost everyone comes to a different result) I know in any case where I find a bright head or phone joker if I ever need one - hats off!



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RiverSong
Legend
--> for task 2 there are meanwhile good and also correct solutions --> check
--> for task 1: only a one-digit cross sum should be formed here.
your results for task 1 are indeed very unusual due to the frequency --> 4 1 8 6 7 9 5
has anyone calculated this in his head? i assume that some of you just made a mistake. that can happen quickly with such a long task.

even if i have calculated my own result 3 times, it doesn't necessarily have to be correct.
had already written in a contribution that we would like to make this a community project
just like this post here if we want to find out the right result together (only for task1). of course I write mine only at the end otherwise some say again --> yours is right and I take that what you have found out not sense of the thing
at the number of likes you see then also whether you simply have fun solving or whether it's just about the stupid psc and some had just quickly slapped something in the hope that it is true and they are there.

i thought of it this way, because the pile of numbers and letters is relatively easy to separate. just take (copy or quote) the list from my next post, calculate it, write your values behind the arrow in each line and put it back in, that way we'll find the right result faster together, ok. by end of next week we'll hopefully have the correct solution to task 1 (without variant, only single digit cross sum. i am also busy elsewhere, that's why the chosen time frame.

my ambition is aroused the learning effect is there for you as well as for me, however, I am not your judge and who has no bock (more) to find out the right result must also not liken.
concerns everything only purely the task 1, the single-digit cross sum, no variant.

please liken first, then copy or quote (next post so that it remains clear), put your values behind the arrow and post until the end of next week, thank you.
sorry for the long text, just can't get it shorter



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RiverSong
Legend



  1. 14338 --&gt

  2. cascade --&gt

  3. 1607 --&gt

  4. waddling duck --&gt

  5. 234

  6. schnurzpiepegal --&gt

  7. 45214 --&gt

  8. as quiet as a mouse --&gt

  9. 2177 --&gt

  10. piston return valve --&gt

  11. 9532 --&gt

  12. schlumpergrete --&gt

  13. 60457 --&gt

  14. lumberjack --&gt

  15. 93 --&gt

  16. flaumenstreicher --&gt

  17. 643 --&gt

  18. hocus pocus --&gt

  19. 8812484 --&gt

  20. luftikus --&gt

  21. 9364527 --&gt

  22. cloud cuckoo land --&gt

  23. 35719 --&gt



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Begbie
Elite
the one-digit sum of the digits is 7.
some people here had also calculated this.
but that wasn't the problem at all, was it?

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