There is a tree having infinite flowers,near the tree there is a magic well, if we dip three flowers in the well 6 flowers will come out if we put ten flowers 20 will come out just like addition 10+same 10= 20, near the well are three graves,now we have to put equal amount of flowers on the three graves, we have to dip how many flowers we take from the tree before putting them on each graves.
NOTE:no flower should be left in the final.
NOTE:you can only pluck how many flowers from the tree but only one time.
Main Question:how many flowers should you pluck from the tree at only 1 time.
Thank you Danish for your submission!
November 8th, 2015 | Category: 
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I may not understand, but I’d pick only 1 single flower from this wonderful tree. I’d dip the flower in the magical fountain, then I’d have 2 flowers. I’d place one flower on a grave and dip the other. I now have 1 flower on a grave & 2 flowers in my hand. I place these on the remaining 2 graves.
I think this satisfies the riddle.
I was thinking of plucking 1/2 of the flowers from the tree (1/2 of infinity) and dipping them in the well.
Place 1/3 of infinity flowers on each grave.
The tree would still have infinite flowers.
The dead would be suitably honored with infinite flowers on each grave.
I would be infinitely busy – but what a calling.
Sorry, I like both Brent and Crabman’s answers, but it is not Danish’s answer.
Keep trying!
I would pick 15? Then dip them and would have 30..then I would place 10 on each grave.
Honestly, I’m having trouble understanding d the questions
NOTE:you can only pluck how many flowers from the tree but only one time.
Main Question:how many flowers should you pluck from the tree at only 1 time.
For each multiple of 3(y) dipped, you can place y*2 flowers on each grave
3dipped=2 on each grave
6=4 on each grave
9=6
Into infinity
I actually thought about the parameters set by the riddle I found them to be not very clear. One flower picked and two on each grave works for me. So I will give you Danish’s answer, and you will see there is certainly something missing from the question.
7
Reason: we pluck 7 flowers from the tree at once.Now we dip 7 flowers before putting one the first grave now we have 14 flowers now we put 8 flowers on the first grave.How many are left? 6 are left. now we dip 6 flowers again before putting on the second grave. now we have 12 flowers we put 8 of them again on the second grave. How many are left now? 4 are left. now we again dip 4 flowers in the well before putting it on the third grave. now we have 8 flowers. Now again we put 8 of the flowers on the third grave. How many are left? 0.So that is why the answer is 7.
Criteria
1) all flowers plucked, or created from dipping, must be used
2) You can pluck as many flowers as you want from the tree, but you may only pluck once.
3) (in riddle text) an equal number of flowers must be placed on each grave
4) (My inferred intention of the last line of text) All flowers picked from the tree must be dipped.
Main Question: how many flowers should you pluck from the tree, and how do you did them and place on the graves?
This still leaves the riddle writers strategy a bit obscure. Why the complexity rather that picking 3, diping them to have 6 and placing 2 on each grave?
Perhaps additional criteria
5) until the final placement you can place no more that 2/3 of the flowers on a grave
6) you can ably place flowers on one grave before having to dip all remaining flowers
7) once placed on a grave, flowers can’t be added or removed.
I compliment the idea of this riddle.
There is another question (and method) where there is a flower shop opposite the cemetery with the three graves and only after placing the first amount of flowers can you proceed to the magic well. Therefore the number of flowers purchased would be 14. Enter the cemetery and place 8 flowers on the first grave, leaving with 6 to dip into the well. 8 flowers would then be placed on the second grave from the 12, leaving four to be dipped and turned into 8 for the final grave.
This question was presented in a year 10 maths exam.