<s>[INST] Monkeys Dividing Pile of Bananas 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?
I tried bashing this problem out using long algebraic expressions, but I kept messing up and was unable to find a good solution.  Could I get some hints on one?  Thanks!
 [/INST] Let $A,B,C$ be the banana's removed by monkeys $1,2,3$ respectively and $x$ be the number of bananas monkey $3$ has at the end.
Then:
$\frac{1}{8}A+\frac{3}{8}B+\frac{1}{12}C=x$
$\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=2x$
$\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=3x$
Solving the system we get:
$A=\frac{22}{17}x$
$B=\frac{26}{17}x$
$C=\frac{54}{17}x$.
Now, let $x=17k$. 
We need $A=22k$ to be a multiple of $8$ and $4$ (hence $8$)
We need $B=26k$ to be a multiple of $8$ and $4$ (hence $8$)
We need $C=54k$ to be a multiple of $24$.
Hence $k$ must be a multiple of $4$.
So $x$ must be  $17\times 4$.
Therefore the number of bananas is $6\times 17\times 4=408$
And monkeys $1,2,3$ take $88,104,216$ bananas each.
</s>