how to work out the values of integers in specific positions in a number Apologies if this question is a duplicate, but I believe it is not.
If there are three sets of numbers, A, B, and C, and each are integers $1\le n \le9$, occupying the hundreds, tens and unit positions, as follows: ABC; ABC; and ABC,  
and when they are added together, they result in a number BBB. i.e.  
$\overline{ABC}+\overline{ABC}+\overline{ABC}=\overline{BBB}$,
solve for the integers A, B and C?
So far, solutions seem to be amenable to a 'serendipitous' approach e.g. dividing BBB by three ($999/3$; $888/3$ etc.) until the result matches ABC match (as much as possible).
Obviously, this is not a satisfactorily mathematical solution. I would be grateful if a solution to this specific question using algebra could be suggested. 
Also, if the general nature or concept surrounding this problem could be described?
Thank you.
 A: $$300A + 30B + 3C = 100B + 10B + B = 111B \implies 300A + 3C = 81B$$
$$\implies 100A = 27B - C$$
Obviously $B>3$. Beyond that, you'll have to resort a bit of trial and error. I can't think of a way to solve this without that.
A: There are only 9 possible values for $B$.
$BBB=111$ and $BBB=222$ are too small because $BBB/3$ has only two digits.
$333$, $666$ and $999$ do not work, since the tens digit of $BBB/3$ is not $B$ in each case.
In the same way we can also eliminate $555$, $777$ and $888$ since $555/3 = 185$, $777/3 = 259$ and $888/3 = 296$. So we are left with
$3 \times 148 = 444$
I see nothing wrong with a case by case approach when the number of cases is small, as it is here.
A: There's lots of options to solving this; the one you mentioned is indeed one of them.
Another idea: Notice that in the second set of numbers, we get $B + B + B$ $+$ (carry in) $= B$. Here, the (carry in) number can only be $0$, $1$, or $2$ since we're only adding three numbers. I think you can then easily work out that $B$ can only be $4$, $5$, or $9$. You're left with $444$, $555$ or $999$. That reduces the number of options, and at this point you can just divide by three to check.
