a) $101,111,121,202,212,222$.

b) I. If a number meets the stated requirements, then:

Since the number is divisible by 6, and thus also by 2, its unit digit must be even, i.e., either 2 or 8. Therefore, only the numbers $222,252,282,828,858,888$ (1) can meet the requirements.

II. These numbers meet the requirements; they are even, and their digit sums are $6,9,12,18,21,24$, which are all divisible by 3; thus, these numbers are divisible by 3 and therefore also by 6. From I. and II., exactly the numbers (1) meet the requirements.

c) The hundreds and units digit of a three-digit symmetric number is one of the nine digits 1, 2, ..., 9, and the tens digit is one of the ten digits $0,1,2, \ldots, 9$. Therefore, the total number of such numbers is $9 \cdot 10=90$.

d) Each of the hundreds and units digits 1, 2, ..., 9 appears exactly 10 times in the numbers just listed, and each of the tens digits $0,1,2, \ldots, 9$ appears exactly 9 times.

Therefore, the sum of the units digits contributes

$$
10 \cdot(1+2+\ldots+9)=450
$$

the sum of the tens digits, each multiplied by 10, contributes

$$
9 \cdot 10(0+1+2+\ldots+9)=4050
$$

the sum of the hundreds digits, each multiplied by 100, contributes

$$
10 \cdot 100(1+2+\ldots+9)=45000
$$

Thus, the total sum sought is $450+4050+45000=49500$.
## Task 2 - 340812

A three-digit natural number is called "symmetric" if and only if its hundreds digit is equal to its units digit.

a) Form all those three-digit symmetric numbers in which only the digits $0,1,2$ appear (each possibly multiple times or not at all)! No justification is required.

b) Form all those three-digit symmetric numbers that are divisible by 6 and in which only the digits 2, 5, 8 appear! Prove that exactly the numbers you specify meet the given conditions!

c) Determine the number of all three-digit symmetric numbers!

d) Determine the sum of all three-digit symmetric numbers!