priest has some flowers with him.and there are 3 temples and 3 ponds before them.he has to wet the flowers before he put inside the temple. he put the same no of flowers at each temple.pond has feature that is it doubles the flowers which are shrink with water.finally he does not have any flowers.how many flowers are there with the priest intially at minimum.

• ans:7
  initially he has x flowers and assume he put y flowers in every temple
  1).x->2x;and(2x-y)left
  2).(2x-y)->(4x-2y);and(4x-3y)left
  3).(4x-3y)->(8x-6y);and(8x-7y)left but it is zero;so 8x=7y the min value of integer x to satisfy this eq is:7
• 13 years agoHelpfull: Yes(5) No(1)
• initially pond temple left
  x 2x y 2x-y
  2x-y 2(2x-y) y 4x-3y
  4x-3y 2(4x-3y) y 8x-7y
  
  finally he doesnot left wid any flower. so 8x-7y=0
  8x=7y
  x=7y/8
  now flowers cant be in fraction, so y=8 and x=7. hence initially he had 7 flowers
• 13 years agoHelpfull: Yes(4) No(0)
• answer is 7.
• 13 years agoHelpfull: Yes(2) No(1)
• 7 flowers
  Let initial no. Of flower be F
  After putting in water , now he has 2F flowers
  Let Y be the no. Of flowers put insite the temple.
  He is left with 2F-Y flowers
  Again, putting in water, he has now 2*(2f-y) flowers
  =4f-2y..
  Puting y flowers in second temple he is left with 4f-3y flowers
  Following same procedure third time he will left with 8 f -7y flowers, and this will be equal to 0 as in the end he has left with no flowers
  So (8f-7y)=0
  This eq will be valid only if f=7and y=8
  Hence he initially has 7 flowers and at every temple he put 8 flowers
• 9 years agoHelpfull: Yes(0) No(0)
• ans:7
  Initially he has 7 flowers.
  After shrink with water it will become 14... At 1st temple he will keep 8,then he has left with 6.
  Then after shrink with with water it will become 12... At 2nd temple he will keep 8, then he will left with 4.
  Then after shrink with water it will become 8... At 3rd temple he will keep 8, then he will left with 0.
  
  
• 8 years agoHelpfull: Yes(0) No(0)