Small number problem

Since the required number $n$ is equal to three times its sum of digits, $n$ must be a multiple of $3$ . As single-digit $n$ is equal to its sum of digits, there is no solution, when $n$ is single-digit.

For two-digit $n$ , let it be $10a+b$ , where $a \ge 1$ and $b \ge 0$ are single-digit integers. Then, $10a+b = 3(a+b) \implies 7a = 2b$ . Since $2$ and $7$ are primes, there is only one solution, that is $a = 2$ and $b = 7$ .

For three-digit $n$ , the smallest possible $n$ or multiple of $3$ is $102$ . But the largest possible sum of digits is $9+9+9 = 27$ and three times it is $81$ , which is less than $102$ . Therefore there is no solution for three-digit $n$ . Same argument for four or more digit $n$ . Hence we have only $\boxed 1$ solution.

There are no answers with one digit.

For a two digit number $\overline{AB}$ , we need to solve $10A+B=3A+3B$ , which becomes $7A=2B$ . The only non-zero solution to this is $27$ .

For three digit numbers, note that largest possible digit sum is $27$ , and three times this is $81$ ; but the smallest three digit number is $100$ , so there are no solutions.

This argument is easily extended for any greater number of digits; so there is just $\boxed1$ solution.