$x$, $y$, $z$ are distinct non-zero digits such that $zyx+zyx+zyx=xxx$ (concatenation, not multiplication) ...

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You're given an integer p with 3 places, such that $p=(100z+10y+1x)$ with $z,y,x$ all different from each other, greater than zero and the following info:
$\quad z\quad y\quad x$
$+z\quad y\quad x$
$\quad z\quad y\quad x$
$-----$
$\quad x\quad x\quad x$

Which means:
$$111x=3(100z+10y+1x)$$
Find p.

So,
i) I tried to use a system of equations in order to solve the problem, but I don't know what I'm doing wrong. $$3x=x+a \Rightarrow a=2x \tag{1}$$ $$30y+2x=10x+b\tag{2}\Rightarrow 30y-b=8x$$ $$300z+10x+b=100x\tag{3}$$
However, I have 4 variables and 3 equations which made me think that either I'm doing something really stupid or this approach (using the basic sum) is incorrect. What's wrong with this attempt?

ii) Is there an area in number theory that studies more abstract versions of this puzzle? For instance:
Given an integer $p$, with $p > 0$, with $k$ places, such that:
$$p=10^k\cdot x_1+10^{k-1}\cdot x_2 + 10^{k-2}\cdot x_3+...+10x_{k-1}$$ find an integer $\lambda > 0$ such that
$$\lambda p=10^k\cdot x_m+10^{k-1}\cdot x_m + 10^{k-2}\cdot x_m+...+10\cdot x_m$$
with $m\in (1,2,...,k)$.

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0
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All of the comments hint at this, and even work it out with reference to modular arithmetic, but the easiest way to see matters is that three times a number ending in $x$ ends in $x$. The only possible digits with that behavior are $0,5$ (if this is not immediately evident, try multiplying each digit by $3$ and looking at the one's place digit of the product). Since $000$ is not a positive number, $xxx=555$. The starting numbers are then identified as $\frac{555}{3}=185=zyx$.

5
On

We obtain $$37x=100z+10y+x$$ or $$36x=100z+10y,$$ which gives that $x$ is divisible by $5$ and since $x\neq0$, we obtain $x=5$ and $$10z+y=18.$$ Can you end it now?