Case enumeration $\quad]$  [ Classical inequalities (other) ]

Find all numbers that are 12 times the sum of their digits.

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Let the desired number contain $n$ digits. Then it is not less than $10^{n-1}$ (this is the smallest $n$-digit number). On the other hand, the sum of its digits is not more than $9 n$ (since each digit is not more than 9). Therefore, the desired number does not exceed $12 \cdot 9 n=108 n$. But according to Bernoulli's inequality (see problem $\underline{30899}$)

$10^{n-1}=1000 \cdot 5 \cdot 10^{n-4}>200 \cdot 5 \cdot(1+9(n-4))=200(45 n-175)>200 n$ for $n \geq 4$.

The desired number obviously cannot be a single-digit number. It remains to consider two cases.

1) $n=2$. But $10 a+b<12(a+b)$ ( $a$ and $b$ are digits).
2) $n=3$. In this case, we get $100 a+10 b+c=12(a+b+c)$, from which $11(8 a-c)=2 b$, i.e., $b$ is divisible by 11. Therefore, $b=0,8 a=c$, from which $a=1, c=8$.

## Answer

108.