Three men possess a pile of money. One man owns 1/2 of the pile, another one owns 1/3 and the third man owns 1/6 of the pile. Each man takes some money from the pile, until nothing is left.
But then the first man returns
1/2 of what he took, the second man returns 1/3 of what he took, and the third man returns 1/6 of what he took.
If the returned total is divided evenly among the men, it is found that each then has what he is entitled to. How much money was in the original pile, and how much money did each man take?

Question

Three men possess a pile of money. One man owns 1/2 of the pile, another one owns 1/3 and the third man owns 1/6 of the pile. Each man takes some money from the pile, until nothing is left. 
                              But then the first man returns
1/2 of what he took, the second man returns 1/3 of what he took, and the third man returns 1/6 of what he took.
If the returned total is divided evenly among the men, it is found that each then has what he is entitled to. How much money was in the original pile, and how much money did each man take?

jack1 · Answer

$pile = x$
$a + b + c = x$
$ \large \frac 36a + \frac 26b + \frac 16 c = y $
so at this stage, the pile in the middle is y, and 
man 1 has 3/6 of what he took initially
man 2 has 3/6 of what he took initially
man 3 has 5/6 of what he took initially
... with me so far?

jack1 · Answer

sorry, typo
$pile = x$
$a + b + c = x$
$ \large \frac 36a + \frac 26b + \frac 16 c = y $
so at this stage, the pile in the middle is y, and 
man 1 has 3/6 of what he took initially
man 2 has 4/6 of what he took initially
man 3 has 5/6 of what he took initially
... with me so far?

jack1 · Answer

pile y divided evenly between the 3 guys:
$\large man~ 1~ now~ has~ \frac 36~ a + \frac y3 = \frac 36 x$
$\large man~ 2 ~now~ has~ \frac 46 b + \frac y3 = \frac 26 x$
$\large man~ 3 ~now~ has~ \frac 56 c + \frac y3 = \frac 16 x $

jack1 · Answer

are you able to rearrange these equations to solve? @dlearner

anonymous · Answer

$$g(x)=\sqrt[3]{-x-1}$$

jack1 · Answer

?

jack1 · Answer

solve using system of simultaneous equations, and get all answers in terms on one of the constants (a,b,or c)
then assume a value for the constant that works
ie I chose "c"
a = 33c
b = 13c
x = 47c
y = 21c

if c = 6, then
x = 282
y = 126
a = 198
b = 78
c = 6

man 1 is entitled to 3/6 of x = 141
man 2 is entitled to 2/6 of x = 188
man 3 is entitled to 1/6 of x = 47

jack1 · Answer

sorry... man 2 = 94

jack1 · Answer

my bad

anonymous · Answer

Hello jack... i think you misunderstand the question...when they take money it is not necessarily what they own. do you follow?

jack1 · Answer

ah, ok
then your answers are a b and c
a is what man 1 took initially (198)
b is what man 2 took initially (78)
and c is what man 3 took initially (6)

however each man then put money back, and split that pile evenly, and ended up with what they were entitled to (man 1 = 141, man2 = 94, man 3 = 47)
follow?

anonymous · Answer

gimme a minute.... I think i'm the one who is misunderstanding the question :)

jack1 · Answer

yep ;)

anonymous · Answer

thanks man  :)

jack1 · Answer

all good bro

perl · Answer

*