Find all natrual numbers that are $13$ times bigger than the sum of their digits. 
Find all natural numbers that are $13$ times bigger than the sum of their digits.


I had a solution and just wanted to verify it. By solving the equation $$13(a_1+a_2+a_3+\dots a_n)=\overline{a_1a_2\dots a_n},$$ we can get $n=3$ and by putting the number $\overline{abc}$ in the equation, I got three answers.
Right? Thanks!

An $n$-digit number is  $\ge 10^{n-1}$, but the sum of its digits $\le 9n$. This give us the inequality $10^{n-1}\le 13\cdot 9n=117n$, which easily leads to $n\le 3$. 
Then $13\cdot(a+b+c)=100a+10b+c$ leads to $87a=3b+12c$, or $29a=b+4c$. 


*

*With $a=0$, we arrive at $b=c=0$.

*With $a=1$, we arrive at $b+4c=29$, hence $5\le c\le 7$, with each case leading to a valid solution

*With $a\ge 2$, $9+4\cdot 9\ge b+4c\ge 58$ leads to a contradiction


That's four solutions in total if we allow $0$ a a solution
