There are 4 parallel rivers and 3 temples.
Each of the temples is between 2 rivers flowing parallel to each other. (R T R T R T R)
To visit the temple the worshiper has to cross the river each 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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We can solve the riddle using trial and error method.
Since we have to find the minimum number we can start with 1 flower and see if the conditions are met, if not then we go to next number 2, so on and so forth.
After some trial you will get the answer at 7 flowers.
If he has 7 flowers initially, and if he offers 8 flowers at each temple then all the conditions will be met.
Initial Flowers = 7.
After crossing first river he has = 14 flowers.
1st temple he offers = 8 flowers.
Flowers left after offering = 6 flowers.
After crossing second river he has = 12 flowers.
2nd Temple he offers = 8 flowers.
Flowers left with him = 4.
After crossing third river he has = 8 flowers.
3rd temple he offers = 8 flowers.
Flowers left with him = 0.
To mathematically arrive at this solution, you can go through the steps below.
Let’s assume that he offers x flowers at every temple.
Thus, he must have only x flowers left at the end of 3rd temple, otherwise he will not have zero flowers at the end of Fourth River.
Also we can say that after offering flowers at second temple he should be left with x/2 flowers. (Since after crossing the river they would double to x)
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.