Critical Integers

Given an $n$ -digit number $k$ , 12 times the sum of $k$ 's digits is as most $12\cdot 9n=108n$ . If $k$ is equal to 12 times the sum of $k$ 's digits, then $k\le 108n$ . We have $n=\lfloor{\log_{10}(k)}$ , so we can solve this inequality to find $n\le 3$ .

• Certainly $n\neq 1$ , since there are no 1-digit multiples of $12$ .
• If $n=2$ , then $k=10\cdot a+b=12\cdot(a+b)$ for some digits $a,b$ . But then $a+b=0,$ so $a=0,b=0$ , which doesn't yield a positive $k$ .
• If $n=3$ , then $k=100\cdot a+10\cdot b+c=12\cdot(a+b+c)$ for some digits $a,b$ . Solving this system gives $a=x+y,b=11x,c=6x+8y$ for some $x,y$ . The only choice of $x,y$ that yield digits $a,b,c$ are $x=0,y=1$ , yielding $a=1,b=0,c=8$ , so $k=108$ .

Therefore, the only number equal to 12 times the sum of its digits is $108$ , so the answer is $\boxed{1}$ .