Every term of a sequence, starting from the second, is equal to the sum of the previous term and the sum of its digits. The first elements of the sequence are

$$
1,2,4,8,16,23,28,38,49, \ldots
$$

Is it possible that 793210041 belongs to this sequence?
We know that a number and the sum of its digits leave the same remainder when divided by 3. In each case, if the number leaves a remainder of 1 when divided by 3, then the number plus the sum of its digits leaves a remainder of 2 when divided by 3, and if the number leaves a remainder of 2, then the sum of the number and the sum of its digits leaves a remainder of 1 because $2+2=4$ leaves a remainder of 1.

Calculating the remainders of the sequence when divided by 3, we obtain a new sequence

$$
1,2,1,2,1, \ldots
$$

that is, a periodic sequence where only the remainders 1 and 2 appear. Since the number 793210041 is divisible by 3, it does not belong to the sequence.