# 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!

• @kingW3 I've got the answers $195,117,156$. – Taha Akbari May 11 '17 at 10:55
• You're right I've started counting from $13\cdot 10$, seems correct since trivially there are no solutions past $13\cdot 27$ (and $13\cdot 27$ is not a solution). – kingW3 May 11 '17 at 10:56
• I got the same results. I don't know you approach, but my approach leads me to the equation $b+4c=29a$ with $a,b,c \in \{0, \dotsc, 9 \}$. Obviously this enforces $a=1$, and then you get precisely those three solutions., – MooS May 11 '17 at 10:58
• Since this is an Q&A site, you should post an answer to your own question (and of course provide some insights on how to arrive at $n=3$ etc.). – MooS May 11 '17 at 11:00

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