
fibonacci puzzle problem
Three men possess a pile of money, their shares being ½, 1/3, 1/6. Each man takes some money from the pile until nothing is left. The first returns ½ of what he took, the second 1/3, and the third 1/6. When the total so returned is divided equally among the men, it is found that each posses what he is entitled. How much money was in the original pile and how much did each man take from the pile?

Hello, lauren2988!
I did not come up with a unique solution . . .
Let the three men be: .
. . And they took dollars, respectively.
returned half of his money.
. . He kept and returned dollars.
returned a third of his money.
. . He kept and returned dollars.
returned a sixth of his money.
. . He kept and returned dollars.
The total returned is: . dollars.
This was divided equally and returned to the men.
. . They each had: . dollars returned.
has: . . which is equal to his original share: .
. . This simplifies to: . .[1]
has: . . which is equal to his original share: .
. . This simplifies to: . .[2]
has: . .which is equal to his original share: .
. . This simplifies to: . .[3]
From [3], we have: .
Substitute into [1] and [2] and both of them become: .
Hence, there is no unique solution.
We have: .
On the right, replace with a parameter
. . and we have: .
The total amount of money is: . dollars.
Since the original amount was divisible by 6, must be a multiple of 6: .
And the parametric equations become: .
The total amount of money is: . dollars ... for any positive integer