Discussion: Christmas Contest 2018

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    Your Ikariam Team

    • About question 6, I found the word chaos in the sixth column from the left, starting from row three and reading downwards. I wasn’t sure if it really was an Ikariam word until I found the description of the Colossus miracle “… ensures chaotic escape and confusion amongst your opponents.” In other words, it causes chaos among your opponents. So, I think “chaos” should be accepted as an Ikariam related word, and people who found it should be rewarded, not penalised for doing so.
    • Evening All Ikariam rulers

      Hope you enjoyed this 2018 Christmas Contest, and I can only guess that the grandma song has been ringing in your heads for the past 2 days :phatgrin:

      I have now published the answers to that question; I apologise for the delay, but I had to re-check all the answers/scores again to every question to make sure everyone has been 'fairly judged'.
      The final list will (should) be out in less than 48 hours.

      Wishing everyone a Merry Christmas and a Happy Holiday.
      A Festive Season To All!!
    • Evening Dot!

      Thank you for all your hard work in organizing and running this year's Christmas contest! Traditions are important.
      Wishing you Merry Christmas and may the New Year bring you health, happiness and peace!

      Cass,
      =^_^=

      "On the contrary, woman is the best equipped machine ever went to battle"
      -Richard Le Gallienne
      "Asking polite with a gun in your hand is better than asking polite with nothing."
    • Cassiopea wrote:

      Happy New Year to all! In spirit of the contest...what's the answer? You will like it, I promise! ;) FB_IMG_1546552091907.jpg
      We can easily guess the answer is 2019, but to justify this we need to figure out what the above figures mean.
      We know that all primitive(i.e. gcd of all three =1) Pythagorean triples are of the form (u^2-v^2, 2uv, u^2+v^2), where u^2+v^2 is the hypotenuse. Clearly 2019 cannot be of the form u^2+v^2 since
      2019≡3 (mod 4), so the triple in question is not primitive.
      Now 2019=3*673 and 3 again cannot be of the form u^2+v^2, but 673=12^2+23^2. So take u=23, v=12. u^2-v^2=385, 2uv=552. (385, 552, 673) is a primitive Pythagorean triple. Then we just triple the triple and get (1155, 1656, 2019) and plug those back and check


      After hours of trial and error I have figured out star=1, tree=6, sock=5, and so the answer is (1656^2+1155^2)^0.5=2019