Answer this puzzle?

Replies

• 

  Morningdot Hablu

  On the first one 0,1,2,3,5,7 and on the 2nd one 0,1,2,4,6,8.
  
  Am i correct??
• 

  vinsocorp

  Mohit its a good try. think how u show 9,19,29.one clue: one digit number u can show it by single cube,another cube is eliminated.
• 

  silverscorpion

  Well, there are atmost 31 days in a month. So, till 31, there are 2 numbers which have repeated digits - 11 and 22.
  
  So, both the cubes must have 1 and 2, or they cant display 11 and 22.
  
  So, there ia 1 and 2 in both cubes. There are 8 digits remaining (3 4 5 6 7 8 9 0)
  There are eight sides remaining in the two cubes, totally..
  
  Hence, the remaining 8 digits can be filled in any manner in the remaining sides..
  
  For example,
  1 2 3 4 5 6 for the first cube,
  1 2 7 8 9 0 for the second is a possibility.
  
  Or, 1 2 4 6 8 0 for first,
  1 2 3 5 7 9 for second
  
  etc.. Any combination of the above will be fine, as long as both cubes have 1 and 2..
• 

  Morningdot Hablu

  vinsocorp

  Mohit its a good try. think how u show 9,19,29.

  For 9
  0 and reverse of 6
  for 19
  1 and reverse of 6
  for 29
  2 and reverse of 6.
  .....
  

  vinsocorp

  one clue: one digit number u can show it by single cube,another cube is eliminated.

  You doesn't mentioned this in question.
  And i tell you that it's a vary simle question when you alow for the above quotes.
  To make a question standard you have to reduce the probablity of finding a answer.
• 

  vinsocorp

  Mohit u r very well in solving puzzles.here after i maintain standards in my question. ......what u doing...
• 

  marustech

  very good .. but how we get 7,8,9 separateiy
• 

  marustech

  great thinking
• 

  learner

  we can also do this sequence
  1st cube: 1,2,3,6,4 2nd cube: 0,1,2,5,7,8
• 

  Saandeep Sreerambatla

  silverscorpion

  Well, there are atmost 31 days in a month. So, till 31, there are 2 numbers which have repeated digits - 11 and 22.
  
  So, both the cubes must have 1 and 2, or they cant display 11 and 22.
  
  So, there ia 1 and 2 in both cubes. There are 8 digits remaining (3 4 5 6 7 8 9 0)
  There are eight sides remaining in the two cubes, totally..
  
  Hence, the remaining 8 digits can be filled in any manner in the remaining sides..
  
  For example,
  1 2 3 4 5 6 for the first cube,
  1 2 7 8 9 0 for the second is a possibility.
  
  Or, 1 2 4 6 8 0 for first,
  1 2 3 5 7 9 for second
  
  etc.. Any combination of the above will be fine, as long as both cubes have 1 and 2..

  
  This is the straight forward simple answer!!
  
  😀