Jump to content
BrainDen.com - Brain Teasers


  • Posts

  • Joined

  • Last visited

  • Days Won


wolfgang last won the day on April 27 2017

wolfgang had the most liked content!

About wolfgang

  • Birthday 12/25/1953

Profile Fields

  • Gender
  • Interests
    Chess,Logical thinking

Recent Profile Visitors

The recent visitors block is disabled and is not being shown to other users.

wolfgang's Achievements

  1. e.g.( As the numbers are from 1 to 6 ) can be written like this: 1126621531323323123152261152311452242363622464
  2. As the numbers are from 1 to 6 that means it may have a relation to dice ( by using two dice): 11 means A 12 means B 21 means C something like that...
  3. It has something to do with names, but I have no idea.
  4. The common number between 30 and 18 is 90. So we bundle the sausages together and cut them into 3 equal parts, resulting 90 pieces. 90/18= 5 pieces for each person.
  5. (BR / BL +UR) UL Thus... (8/4 +2) 3= 12 (2/1 + 6) 4= 32 (9/3 +4) 2= 14 and for the last one.. (6/2 +3) 5 = 30
  6. The Big Bang theory is the prevailing cosmological model for the universe from the erliest known periods through its subsequent large-scale evolution.The model accounts for the fact that the universe expanded from a very high density and high temperature state Some estimates place this moment at approximately 13.8 billion years ago, which is thus considered the (age of the universe) After the initial expansion, the universe cooled sufficiently to allow the formation of subatomic particles, and later simple atoms. Giant clouds of these primordial elements later coalesced through gravity to form stars and galaxies. My question was so: what if the whole universe was inside a colsed room? and the galaxies were just like dust particles hanging there , if these paricles are subjected to a continuous diverging gravity ,so they will expand. That means , that there was NO start point and NO Big bang theory ! What do you think?
  7. We should devide the 12 coins into three groups, 5,5,and 2 1- for the first five, we weigh 2 against 2, leaving one coin aside. a- if they are blanced,we put them aside. b.- if they areĀ“nt balanced. we should weigh the heavier two with any other two coins :- now... if they balanced so the counterfeit coin is one of the two light coins,which will be easy to be find with the third weigh.....and if they are heavier,then we should weigh them one against one,,,and they heavier is the counterfeit coin. 2- for the 2nd five .we will do the same 2 against 2,leaving one coin aside:- a- if they are balanced, so one of the two coins left aside is the counterfeit coin, which will be easy to find with the 3rd weigh. b- if they are not balanced....see above.
  8. Amasing!!! you did find all the sequences from 0 to 7 ...but 5 is wrong..
  9. No..sorry...try again Note that: abcdef= 0 Note that: abcdef= 0
  10. It seems to be hard....so let me give you a hint....which is: 5
  11. abcdef bc abged abgcd fgbc .........? afedcg abc
  • Create New...