There are three rivers and after each river lies a grave. A man wants to leave the same number of flowers at each grave and be left with none at the end. However, each time he passes through a river, the number of flowers he has doubles. How many flowers does he have to start with so that he is left with none at the end? And how many does he leave at each grave?

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Actually Geoff – he would start with 7 rather than 14 since the 7 will double before he reaches the first grave. 7 becomes 14 – 8 = 6 becomes 12 – 8 = 4 becomes 8.

Close enough!

You did, however, leave out the first step:

You start with 7 flowers. After crossing the first river you now have 14, you leave eight.

After crossing the second river (you started with 6) you now have 12, you leave eight.

After crossing the third river (you started with 4) you now have eight, you leave ’em.

Nicely done, Geoff,

You’re today’s winner.

This problem has an infinite number of solutions modeled by the equation 8a=7n, where “a” is the amount of flowers the man starts with and “n” is the number of flowers he leaves at each grave. The simplest and possibly trivial solution would be to start with 0 flowers and leave 0 flowers at each grave. A more significant solution would be to start with 7 flowers and leave 8 at each grave. Any positive integer multiple of this solution also satisfies the conditions. For example, the man starts with 14 flowers and leaves 16 at each grave; so, 14 doubles to 28, and 28-16= 12; 12 doubles to 24, and 24-16= 8; 8 doubles to 16, and 16-16= 0. The result is the same if the man starts with 21 flowers and leaves 24 flowers at each grave, or starts with 28 and leaves 32, etc.

Thank you Daniel for further clarification.

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.

What would be the answer if there were 7 rivers. Please reply fast.

7 flowers

7 doubles to 14 and then 8 flowers should be placed on each grave. You r left with 6 flowers then passing through the river it becomes 12, then again place 8 on next grave. You are left with 4 flowers, passing through the river 8 and then third grave too gets 8 flowers.