Temple Logic Problem!

Level pending

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.Then the minimum number of flowers he had before entering the first temple be ( A ) and number of flowers he offered at each temple be ( B ). Then what is the value of ( A+B )?

This is not original problem.


The answer is 15.

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1 solution

Maggie Miller
Jul 17, 2015

After entering the third temple, the man has B B flowers. After entering the second temple, the man has 1 2 B + B = 3 2 B \frac{1}{2}B+B=\frac{3}{2}B flowers. After entering the first temple, the man has 1 2 ( 3 2 B ) + B = 7 4 B \frac{1}{2}(\frac{3}{2}B)+B=\frac{7}{4}B flowers. Therefore, B B is the smallest value so that B , 3 2 B , 7 4 B B,\frac{3}{2}B,\frac{7}{4}B are even integers, so B = 8 B=8 .

Then before entering the temple, the man had 1 2 7 4 8 = 7 \frac{1}{2}\cdot \frac{7}{4}\cdot 8=7 flowers, so A = 7 A=7 .

Thus, the answer is 7 + 8 = 15 7+8=\boxed{15} .

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