A3. How many natural numbers less than 1000 have a sum of digits divisible by 7 and are multiples of the number 3?
(A) 7
(B) 19
(C) 21
(D) 28
(E) 37
A3. Multiples of the number 3 have a sum of digits divisible by 3. The sum of the digits of the sought numbers is thus divisible by 21. Since the sum of the digits of each number less than 1000 is less than or equal to $9+9+9=27$, and the sum of the digits of the sought numbers is a multiple of 21, the only possibility is that this sum is equal to 21.

The smallest digit is at least $21-18=3$ and the largest is $\frac{21}{3}=7$. All possible combinations of digits are the triples $(3,9,9),(4,8,9),(5,7,9),(5,8,8),(6,6,9),(6,7,8)$, and $(7,7,7)$. A triple with three different digits determines six possible numbers, a triple with two identical digits determines three possible numbers, and the triple $(7,7,7)$ determines one number. There are a total of $3 \cdot 6 + 3 \cdot 3 + 1 = 28$ such numbers.