Flowers! Flowers!

There are 3 rivers and after each river lies a grave. So there are 3 rivers and 3 graves. A man wants to leave the SAME amount of flowers at each grave, and be left with none at the end. What happens though is that each time he passes through one of the rivers the number of flowers he has doubles. So he has to start off with what number of flowers, taking into consideration that they double, so that he is left with no flowers whatsoever at the end?

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Flowers! Flowers! Solution

Let f be the starting number of flowers, and g be the number left at every grave.

((f*2-g)*2-g)*2-g=0
(4f-2g-g)*2-g=0
8f-4g-2g-g=0
8f-7g=0
8f=7g

Assume f and g are nonnegative integers.

f=7g/8, so 7g must be a multiple of 8, so g must be a multiple of 8.

g=8f/7, so 8f must be a multiple of 7, so f must be a multiple of 7.

So all nonnegative g that are multiples of 8 are solutions. The first few are listed:

f=0, g=0.
f=7, g=8.
f=14, g=16, etc.

Of these, the simplest solution is to start off with zero flowers, and snub each grave.

Oh, No, Another One of Those ''Year'' Puzzles

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