<s>[INST] Beads and necklaces We have a several number of beads in a box and want to make 3 necklaces. 
We firstly take some out of the box, put $3/4$ of them to the 1st necklace and equally share the rest $1/4$ to the 2nd and 3rd. 
Then we take some more out of the box, put $1/4$ of them for the 2nd necklace and equally share the rest $3/4$ to the 1st and 3rd. 
Finally we take the remaining beads, put $1/12$ to the 3rd necklace and equally share the rest 11/12 to the 1st and 2nd. 
We then count the beads of the 3 necklaces and find them to be in ratio $3:2:1$. What is the least number of beads that we initially had in the box?

$A$ = number of beads initially in the box
We firstly take $x$, put $3x/4$ in the 1st necklace and equally share the rest $x/4$.
So the 1st necklace has $3x/4$ 
2nd  has $x/8$
3rd has $x/8$
Then we take $y$, put $y/4$ in the 2nd necklace and equally share the rest $3y/4$
So 1st now has $3x/4+3y/8$
2nd  has $x/8+y/4$
3rd has $x/8+3y/8$
Then we take the remaining, put $(Α-x-y)/12$ in the 3rd necklace and equally share the rest $11(A-x-y)/12$ in the other two.
So the 1st necklace has \begin{align}3x/4+3y/8 + 11(Α-x-y)/24 =&\; 3x/4+3y/8-11x/24-11y/24+11Α/24\\ =&\; 7x/24-2y/24+11Α/24\end{align}
2nd  has  $$x/8+y/4+ 11(Α-x-y)/24 = 11Α/24-8x/24-5y/24$$
3rd has  $$x/8+3y/8 +(Α-x-y)/12 = Α/12+x/24+7y/24$$
And we know that $$(7x/24-2y/24+11Α/24) : (11Α/24-8x/24-5y/24) : (Α/12+x/24+7y/24) = 3:2:1$$
Any ideas on how to continue?
 [/INST] The last result implies two equations:
\begin{align}
\frac{\frac{11}{24}Α+\frac{7}{24}x-\frac{2}{24}y}{\frac{11}{24}Α-\frac{8}{24}x-\frac{5}{24}y}=&\; \frac32,\\[2ex]
\frac{\frac{11}{24}Α-\frac{8}{24}x-\frac{5}{24}y}{\frac{2}{24}Α+\frac{1}{24}x+\frac{7}{24}y}=&\; \frac21.
\end{align}
Solving this system for $x$ and $y$ implies that
$$x=\frac{11}{51}A,\quad\text{and}\quad y=\frac{13}{51}A.$$
We need the following numbers to be integers:
$$A,\;\; x,\;\; y,\;\; \frac{11}{24}Α,\;\;\frac{7}{24}x,\;\;\frac{2}{24}y,\;\;\frac{11}{24}Α-\frac{8}{24}x,\;\;\frac{5}{24}y,\;\;\frac{2}{24}Α,\;\;\frac{1}{24}x,\;\;\text{and}\;\;\frac{7}{24}y.$$
Notice that $51=3*17$, and $24=2^3*3$. Since $11$ is prime, for $x$ to be an integer $A$ must be divisible by $51$. For all other fractions above to be integers it is necessary and sufficient that $A$ also be divisible by $24$. 
Therefore, $$A=\rlap{\underbrace{\phantom{2^3*3}}_{24}}2^3*
     \overbrace{3*17}^{51}=408$$ is the lowest possible value for $A$.
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