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.


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.

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    $\begingroup$ Do you really mean $1 \le n \le 9$ ? [That way allows use of digits 1 and or 9 (or neither)] $\endgroup$ – coffeemath Apr 20 '18 at 11:21
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    $\begingroup$ Seems like an excellent method to me. If you want, write $\overline {ABC}=100A+10B+C$ and $\overline {BBB}=100B+10B+B$ and work from there. $\endgroup$ – lulu Apr 20 '18 at 11:21

$$300A + 30B + 3C = 100B + 10B + B = 111B \implies 300A + 3C = 81B$$

$$\implies 100A = 27B - C$$

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.

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  • $\begingroup$ Considering that this came out of just a whatsapp meme with no serious mathematical intention to it, I tend to agree with you that at this point, without additional data, the numbers A,B or C are unsolvable using algebra (or at least my limited knowledge of it). (I would be grateful if anyone can still solve it algebraicly with this limited information.) And that instead, if the question is given in this form per se, the approach I initially described and elaborated by sedrick and gandalf61 below, is the way to 'solve' it. $\endgroup$ – Abdul-Kareem Abdul-Rahman Apr 20 '18 at 14:16

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.

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  • $\begingroup$ Although I agree with you that in this specific case, this might have been the intention of the initial question-poser, I posed this question with the intention to explore if more general/universal approaches could be derived from an algebraic approach. $\endgroup$ – Abdul-Kareem Abdul-Rahman Apr 20 '18 at 14:19

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.

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