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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The Answer is 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.
Explaination

Explaination:- 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.