A man of faith visits the holy land where three temple are erected in a row offering a surreal divinity to the place. Awestruck at the sight of the magnificent temples, he visits the first temple where he finds that the number of flowers in his basket increases to double the number as soon as he sets his foot inside the temple. Astonished by the miracle he is completely happy and seeks it as a blessing of his god. He offers a few flowers to the stone idol of god and moves out.
Then he visits the second temple, where the same miracle happens again. The number of flowers in his basket again double up in number. With a smile on his face, he offers a few of flowers to the stone idol again and walks out after praying.
The moment he enters the third temple, the number of flowers increases to double again. Now he feels entirely blessed. He offers all the flowers at the feet of the stone idol and walks out in a moment of bliss.
When he walks out, he has no remaining flowers with him. He offered equal number of flowers at each of the temple. What is the minimum number of flowers he had before entering the first temple? How many flowers did he offer at each temple?
Solution:
Let m be initial number of flowers and n be the flowers offered at each temple by the man.
Initial flowers = m
After stepping into the first temple = 2m
Flowers offered at the first temple = n
Remaining flowers = 2m – n
After stepping into the second temple = 4m – 2n
Flowers offered at the second temple = n
Remaining flowers = 2m – n
After stepping into the third temple = 8m – 6n
Flowers offered at the third temple = n
Remaining flowers = 8m – 7n
Now as per the question (8m – 7n) must be equal to 0.
i.e. 8m – 7 n = 0
or 8m = 7y
Now the least numbers required to solve this equation are 7 and 8 for m and n respectively.
Thus m = 7; n = 8
Which means, the man had 7 flowers in the beginning and he offered 8 flowers at each temple.
Submit your Email Address to get latest post directly to your inbox.
