Puzzle:
In a small town, there are three temples in a row and a well in front of each temple. A pilgrim came to the town with certain number of flowers.
Before entering the first temple, he washed all the flowers he had with the water of well. To his surprise, flowers doubled. He offered few flowers to the God in the first temple and moved to the second temple. Here also, before entering the temple he washed the remaining flowers with the water of well. And again his flowers doubled. He offered few flowers to the God in second temple and moved to the third temple. Here also, his flowers doubled after washing them with water. He offered few flowers to the God in third temple.
There were no flowers left when pilgrim came out of third temple and he offered same number of flowers to the God in all three temples. What is the minimum number of flowers the pilgrim had initially? How many flower did he offer to each God?
Solution:
The pilgrim had 7 flowers, initially and he offered 8 flowers to each God.
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Solution:
The pilgrim had 7 flowers, initially and he offered 8 flowers to each God.
Assume that the pilgrim had X flowers initially and he offered Y flowers to each God. From the data in puzzle, there are (8X - 7Y) flowers when the pilgrim came out of the third temple. But it is given that there were no flowers left when he came out of third temple. It means that
(8X - 7Y) = 0 or 8X = 7YThe minimum values of X and Y are 7 and 8 respectively to satisfy above equation. Hence, the pilgrim had 7 flowers and he offered 8 flowers to each God.
In general, the pilgrim had 7N flowers initially and he offered 8N flowers to each God, where N = 1, 2, 3, 4,
8 comments:
Minimum number of flowers with which the piligrim enters the temple = 7
Flowers offered at each temple = 8
let x be the number of flowers
let y be the number of flowers offered at each temple
At temple 1:
Let z be the number of flowers left
thus 2x-y=z
At temple 2:
Number of flowers the piligrim intially has after offering at temple 1= z
Let p be the number of flowers left
thus 2z-y=p
At temple 3:
Number of flowers the piligrim intially has after offering at temple 2= p
After offering at temple 3 he dowesn't have any flowers
thus 2p-y=0
=>p=y/2
thus by solving the above 2 equations we get x=7y/8
As number of flowers should be a digit..y should be a multiple of 8
minimum value for y=8
thus x=7
so Intially the piligrim has 7 flowers and offered 8 flowers at each temple
He had 7 flowers, offered 8 at each temple
compared to Chavali's answer it is teadious and not suitable for long sequence
Anyway let x be initial no. of flowers and y be the flowers offered at each temple.
Then sequence goes as
Initial flowers = x
after crossing first well = 2x
offering @ first temple = y
remaining = 2x-y
after crossing second well = 4x -2y
offering @second temple = y
remaining = 4x-3y
after crossing third well = 8x-6y
offering @ third temple = y
remaining = 8x-7y which is zero.
Therefore, 8x-7y=0;
8x=7y, and least possible solution for this is x=7 & y=8;
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I designed a easier solution. more generalised form.
the answer is
2^n-1 no of flowers intially he was carrying.
2^n no of flowers he kept at first temple.
where n is no of temples.
and 2^n represents 2 power n.
If u think my formula is helpful please post ur comments and follow my blog so that i get to know was that formula useful to anyone...
Patil, what you say is right but I would think a prove is in order. I thought of one.
Lemma 1: For n is the amount of temples visited before we end op with E_n flowers (with E_n = 0 for the final temple), the following holds: 2^n * S - (2^n - 1)I = E_n, where S is the amount of flowers at the start and I is the amount of flowers offered.
Proof by induction
Base case: For 1 temple visited, we double our S and then offer i flowers to the gods. So we have 2S - I = E_1, alternatively, 2^1 * S - (2^1 - 1)I = E_1, so the base case holds for n = 1.
Induction Hypothesis: 2^n * S - (2^n - 1)I = E_n
Induction:
2(E_n) - I = E_n+1
2(2^n * S - (2^n - 1)I) - I = E_n+1.
2^n+1 * S - (2^n+1 - 2)I - I = E_n+1.
2^n+1 * S - (2^n+1 - 1)I = E_n+1.
The lemma has been proven.
We know that if we visit n temples, with S starting flowers and I offered flowers per temple, we have E_n = 2^n * S - (2^n - 1)I (lemma 1). Also we know that E_n = 0 since we end up with 0 flowers at the end.
So we know:
2^n * S - (2^n - 1)I = 0
2^n * S = (2^n - 1)I
Since the coefficients of S and I are one apart, they are per definition co-prime. So except for 1 they have no common divisor. We know a possible integer solution for the equation is S = 2^n - 1 and I = 2^n. Since the numbers are co-prime, we can not S and I by any number (other than the identity element 1) without making the solution non-integer. So S = 2^n - 1 and I = 2^n is the smallest integer solution for the problem.
QED
(I do not know if the part about proving it's the smallest is rigorous enough, anyone know a better way of showing it's the smallest integer solution?)
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