Puzzle Time:
All the rivers are magical, moment a worshiper crosses a river with any number of flowers, it becomes double. (For eg 1 flower becomes 2 flowers, 2 become 4 and so on).
How many minimum number of flowers a worshiper needs 2 carry from beginning such that he offers equal number of flowers in all the three temples, and he is left with zero flowers at the end of fourth river.
Solve this ??
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also we can say that after offering flowers at second temple he should be left with x/2 flowers.
similarly after offering flowers at first temple he should be left with (3x/2)/2 flowers.
Thus he should be carrying (x + 3x/4)/2 flowers initially, i.e. 7x/8 flowers initially.
For number of flowers to be integer, x =8, thus he must carry 7 flowers initially and offer 8 flowers at each temple.
Answer can be 7,14,21etc
And the man gives 8,16,32 flwrs in tmpl
There are so many solutns for dis given prblm
How will u say 7 14 21. Will u explain me
Vaibhav, the question asks for minimum number of flowers. Yes the answers you give are correct but minimum is 7 & 8
7.
7 flowers
7 Flowers & offers 8 Flowers in each Temple.
7 * 2 = 14 (First River)
14-8 = 6 (First Temple)
6*2 = 12 (Second River)
12-8 = 4 (Second Temple)
4*2 = 8 (Third River)
8-8=0 (Third Temple)