The problem is as follows:
$\textrm{Find the sum of a+b+c}$
$\overline{abc}=\overline{c000}_{(3)}$
What I have tried is to transform back to base $10$ the number on base $3$ by doing the following:
$\overline{c000}_{(3)}=3^{3}\times c+3^{2}\times 0+3^{1}\times 0 +3^{1}\times 0 + 3^{0}\times 0$
therefore:
$\overline{c000}_{(3)}=27\times c$
Then I came up with the idea that $c$ can only be (judging from the numeric system above) only $1$ or $2$ since a base $3$ number limits to those values. But if I do multiplication between $1$ or $2$ by $27$. It does not produce a number of three digits.
$27\times 1=27$
$27\times 2=54$
However I know that the sum of the numbers which results from multiplication of a number of divisibility by $9$ (in this case $27$) with any other number is equal to $9$.
Therefore let's say if
$27\times 4=108$
$1+0+8=9$
But the above procedure does not seem to be right as $4$ is off the limits from the base in the number.
I'm not sure if the sum of $a+b+c=9$, but if it is, is there any way to prove it? Did I made a mistake?
If the number $n=\overline{abc}=\overline{c000}_\beta$, then $$ c\beta^3 = 100a+10b+c $$ so $$c(\beta^3-1)=100a+10b$$ with $1\le c < \beta, 0 \le a,b \le 9$. Since $100a+10b<1000$, we conclude that $\beta < 10$. Checking all possibilities then yields $432=2\cdot 6^3$ and $864=4\cdot 6^3$ as the only solutions to the problem.
The base $3$ specification yields no solution. The problem is incorrectly written.