Starting with a natural number, Márcio replaces this number with the sum of its digits, obtaining a new number, with which he repeats the process, until he finally arrives at a number with only one digit. For example, Márcio replaces 1784102 with 23 and then with 8. He also applies this process to lists of $N$ natural numbers, replacing each number in the list with the sum of its digits, thus obtaining a new list of $N$ numbers, with which he repeats the process, until he arrives at a final list of $N$ numbers, each with only one digit.

(a) Starting with $3^{2009}$, what is the final number with only one digit?

(b) Starting with $17^{2009}$, what is the final number with only one digit?

(c) Starting with the list of the first 20092009 natural numbers, does the final list have more digits 4 or 5? How many 9s are there in the final list?
We know that any number and the sum of its digits always leave the same remainder when divided by 9. Thus, Márcio replaces the initial number with another, much smaller, number that has the same remainder when divided by 9, and continues this process until he reaches a single-digit number, which is, of course, equal to the remainder of the division of all the numbers obtained previously, including the first one, by 9. Therefore, what Márcio does is simply a one-step process, which consists of replacing natural numbers with their remainders when divided by 9.

(a) Since $3^{2009}=3^{2008} \times 3=\left(3^{2}\right)^{1004} \times 3=9^{1004} \times 3$, the remainder of the division of $3^{2009}$ by 9 is 0. Therefore, the final number in Márcio's process is 9.

(b) Observe that $17^{2}=(18-1)^{2}=18^{2}-2 \cdot 9+1=$ multiple of $9+1$. Thus,

$$
17^{2008}=\left(17^{2}\right)^{1004}=\text { multiple of } 9+1
$$

and, therefore, $17^{2009}=$ multiple of $9+17=$ multiple of $9+8$. Therefore, the final number in Márcio's process is 8.

(c) Applying the process to the list of natural numbers 1, 2, 3, 4, 5, 6, $7,8,9,10,11,12, \ldots$, the final list is always $1,2,3,4,5,6,7,8,9,1,2,3, \ldots$ Since the remainder of the division of the number 20092009 by 9 is 4, the last number in the final list is 4, and the last six digits of the final list are ..., $8,9,1,2,3,4$. Therefore, this list has the four digits 1, 2, 3, and 4 one more time than the digits 5, 6, 7, 8, and 9. In particular, there are more 4s than 5s in the list. The number of times the digit 9 appears in the list is the number of multiples of 9 that are less than or equal to 20092009. Since 20092005 is the largest multiple of 9 that is less than 20092009, the digit 9 appears $20092005 \div 9=2232445$ times in the list.