12) How many positive integers are equal to three times the sum of their digits? $\begin{array}{lllll}\text { (A) } 0 & \text { (B) } 1 & \text { (C) } 2 & \text { (D) } 4 & \text { (E) infinite. }\end{array}$
12) The answer is (B). Obviously, there are no single-digit numbers, nor numbers with 3 or more digits. The numbers sought will therefore be of the form $10a + b$, with $a$ and $b$ integers and $1 \leq a \leq 9, 0 \leq b \leq 9$. Therefore, we will have $10a + b = 3(a + b)$, from which $7a = 2b$, which has as its only solution that satisfies the given conditions $a = 2, b = 7$. The only number that is equal to three times the sum of its digits is therefore 27.