Numbers $n$ such that $n$ plus the sum of $n$'s digits is $313$ Good morning, everyone. Here is the problem I'm faced with:

The sum of the number $n$ and the digits of $n$ equals $313$. What are the possible values of $n$?

By reasoning I concluded that it has to be a $3$ digit number and by hit and trial I found $305$ as one of the answers. How would I find all possible values of $n$?
It would be great if someone could throw light on how to proceed in a mathematical way and get the values of $n$ (as opposed to brute-forcing/trial-and-error).
I tried as below:
Let our number be $xyz$, where $x,y,z$ are the digits of the number. Then we seek solutions to
$$\underbrace{100x + 10y + z}_{n} + \underbrace{x+y+z}_{n's \; digits} = 313$$
Thus, simplifying, we seek integer solutions to 
$$101 x + 11 y + 2 z = 313$$
I am stuck now as to how to solve this.
Thanks to all in advance.
 A: The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$ 
Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.
A: You are on the right track with:
$$\overline{xyz}+x+y+z=101x+11y+2z=313.$$
$x\ne 1$, because otherwise $11y+2z\le 117<212$.
$x=2$, then:
$$z=\frac{111-11y}{2} \Rightarrow y=9, z=6 \Rightarrow \overline{xyz}=296.$$
A: $$101x+11y+2z=313$$
where $x,y,z$ are numbers between $0$ and $9$.
If $x\le 1$, then $101x+11y + 2z \le 101+11(9)+2(9)=101+13(9)=218 $ 
If $x=2$, then we have $$11y+2z=111$$
Since $z\le 9$, we have $$11y =111-2z \ge 93$$
Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$
Hence $296$ is a solution.
If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.
The solutions are $305$ and $296$.
A: Going from your final conclusion, 
$$101x + 11y + 2z = 313$$ 
it seems sufficient to simply notice that $x,y,z \in \{0,1,2,3,...,9\}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x \in \{1,2\}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know: 
$$x \in \{1,2,3\}$$
From here, it becomes more processes of elimination.

Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.

Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.
Try $y=9$. This yields $z=6$.
Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.

Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.
As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.

Thus, we end up with two solutions:
$$x=3,y=0,z=5 \implies 305 \; \text{ is a solution} $$
$$x=2,y=9,z=6 \implies 296 \; \text{ is a solution} $$
Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to
$$n + s(n) = 313$$
where $s(n)$ denotes the sum of the digits of $n$.
