# 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.

$$101x + 11y + 2z = 313$$

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 $$n$$.

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 $$a=5$$.

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: $$z=\frac{111-11y}{2} \Rightarrow y=9, z=6 \Rightarrow \overline{xyz}=296.$$

$$101x+11y+2z=313$$

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 $$296$$.