# Finding all numbers that are $4$ times their digit sum

A natural number $$n$$ is said to be a good number if and only if $$4$$ times the sum of its digits equals the original number. We have to find out the sum of all such good numbers.

I'm 99% sure the only such good natural numbers are $$12, 24, 36$$ and $$48$$. It can't have one digit, obviously and three digits or more is also not possible. But how do I prove this, as 'obviously' is not a mathematical statement accepted in my class. How do I prove that more than 2 digits is not possible? One digit is unacceptable is easy to prove.

• Might be relevant: oeis.org/A037478 Commented May 10, 2019 at 5:20

Assuming $$n>0$$,
1-digit: $$n=a_1=4a_1 \iff a_1=0 \not >0$$.
2-digit: $$n=\overline{a_2a_1}=10a_2+a_1=4a_2+4a_1 \iff a_1=2a_2 \iff 12, 24, 36, 48$$.
3-digit: $$n=\overline{a_3a_2a_1}=100a_3+10a_2+a_1=4a_3+4a_2+4a_1 \iff a_1=32a_3+2a_2\ge 32$$.
$$m$$-digit: $$n=\overline{a_ma_{m-1}...a_1}=\sum_{i=1}^m a_i\cdot 10^{i-1}=4\sum_{i=1}^ma_i \iff a_1=\sum_{i=2}^m \frac{10^{i-1}-4}{3}a_i \le 9 \iff m\le 2.$$
Hint: Write $$n = \sum_{k=1}^{m}a_k10^k,$$ where $$n$$ is an $$m$$-digits number. Then $$\sum_{k=1}^{m}a_k10^k - \sum_{k=1}^{m}a_k\cdot4 = 0.$$ What happened if $$m \geq 3$$?