AIMEII 2004 #6

[sinkokuyou]

Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio 3:2:1, what is the least possible total for the number of bananas?

For some reason, I got 992...
i sux at this... :(

[sinkokuyou]

This is how I did it, which is wrong, but can someone shows me where I did wrong?
My work:
set 1 as the whole, so 1st monkey took away [unparseable or potentially dangerous latex formula] bananas, and 2nd monkey took away [unparseable or potentially dangerous latex formula] bananas, so the 3rd one got [unparseable or potentially dangerous latex formula].
and set:
1st: [unparseable or potentially dangerous latex formula]
2nd: [unparseable or potentially dangerous latex formula]
3rd: [unparseable or potentially dangerous latex formula]
so according to the given info, we can set up our equations:
[unparseable or potentially dangerous latex formula]
which is really 40a+13b=31
[unparseable or potentially dangerous latex formula]
which is really 8a-b=31
and solve for a and b, I got
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
and since the have to be divisible by 4, so I multiplied 248x4=992, which is not the right answer...

[88bobcat]