<s>[INST] Find all integers $x$ that can't be written as $x=a+b+c$, where $a$ divides $b$ and $b$ divides $c$ with $aFind all integers $x$ that can't be written as $x=a+b+c$, where $a$ divides $b$ and $b$ divides $c$ with $a<b<c$
This implies that $a$ divides $c$.
I can also rewrite $c=zb$ and $b=ya$ so $c=zya$ with $z\neq0$ and $y\neq0$
So $x=a(1+z+zy)$
$z$ cannot be equal to $1$ because $c=zb$ and $b<c$
$y$ cannot be equal to $1$ because $b=ya$ and $a<b$
So I can state that none of the prime number can be write under that form if $a\neq1$.
But what then?
 [/INST] There is no solution for $x=1,2,3,4,5,6,8,12,24$ and solutions for all other $x$.
You should probably consider the case when $a=1$ first. There is a solution to $x-1=z(1+y)$ when $x-1$ is non-prime and bigger than $4$ (since $1+y>2$.)
On to the general equation.
If $X$ has an odd factor $d>5$, then $d-1=2\left(1+\frac{d-3}{2}\right)$, and $ X=\frac{X}{d} + \frac{2X}{d} + \frac{2X(d-3)}{d}$.
So we can reduce the question to $X$ which can be written as $2^n$, $3\cdot 2^n$ and $5\cdot 2^n$.
There is no solution for $1,2,3,4,5,6$.
Since $2\cdot 5-1=3\cdot 3$, we get that $5\cdot 2^n = 2^{n-1}+3\cdot 2^{n-1}+3\cdot 2^{n-1}$.
Since $3\cdot 2^5-1=5\cdot 19$, we have that:
$$3\cdot 2^{n+5}=2^{n}+5\cdot 2^n + 18\cdot 5\cdot 2^n$$
$$x=2^{n+4}=2^n + 5\cot 2^{n+1} + 5\cdot 2^{n+2}$$
Left to check: $x=8,12,24,48$. $48=3(2^4)=3+15+30$.
There is no solution for $8$, since if $c>4$ then $c\geq 6$ and $a+b+c>8$.
For $x=12$, we can't have $a=1$ since $x-1$ is prime. If $a>1$ then $12/a \leq 6$ and $1+y+yz=12/a$ has no solution.
$x=24$ is the last case. For all divisors $a\mid x$, you have either $x/a-1$ is prime, or $x/a-1<6$.
So there is no solution for $x=1,2,3,4,5,6,8,12,24$ and solutions for all other $x$.
</s>