# double the stone puzzle

Story:

I've x number of stones. On my way, I will cross three wells. When I reach a well, then I will throw y number of stones into the well. After that, remaining stones are twice the count....

eg:

If I have 5 stones and I dropped 3 stones into the well. Now remaining 2 stones are twice the count. So I have 4 stones, while leaving the well.

It is continues to all well crossing. At end of travel, I have zero number of stones.

Puzzle:

What is the value of x+y?

Note:

Same number of stones must be dropped into the wells.

Hint:

Solve the puzzle as backwards. Because, there we have a value as zero.

• "After that, remaining stones are twice the count...." ? – Oray Jul 14 '18 at 11:54
• when leaving the well, the remaining stone's count is twice.. – user50721 Jul 14 '18 at 11:56
• twice than $y$ and/or how many stone you throw in a well? – Oray Jul 14 '18 at 11:58
• After throwing y number of stones, then I have 2(x-y) number of stones. Beside, I update my question with an example. Kindly take look on it. – user50721 Jul 14 '18 at 12:03

If I understand the question correctly, every time you throw $y$ amount of stones into a well, at the end the number of stones you have becomes as twice as before, for example let's say we have $x$ amount of stones, if I throw $y$ amount of stones, then I am supposed to have $x-y$ stones left, but magically it becomes $2x-2y$ stones. So what happens if I visit 3 wells and I end up with zero stones after I throw the last $y$ amount of stones.

After revisiting the original question, let's solve it:

Well 1

I had x stones, I threw $y$ stones, I was supposed to have $x-y$ stones then magically I ended up $2x-2y$ stones with $y$ stones in the well.

Well 2

I had $2x-2y$ stones, I threw $y$ stones, I was supposed to have $2x-3y$ stones then magically I ended up $4x-6y$ stones with $y$ stones in the well.

Well 3

I had $4x-6y$ stones, I threw $y$ stones, and I should not have more stones.

As a result

$4x-6y=y$ and $4x=7y$

So the minimum number of stones that you can carry

could be $x=7$ and $y=4$. So $x+y=11$ stones.

if I understand it correctly of course.

• what a man....!! – user50721 Jul 14 '18 at 12:19