Solution

a) First, we multiply the digits of 79, obtaining $7 \times 9=63$, and then we sum the digits of this product, obtaining $6+3=9$. Therefore, the transformed number of 79 is 9.

b) Claudia's game has two stages: the first, in which she multiplies the digits, and the second, in which she sums the digits of the product found, in the case where this product has more than one digit. For 3 to be obtained as the transformed number of a number in the first stage, this number can only be 13 or 31. For 3 to be obtained as the transformed number of a number in the second stage, the result of the first stage must be a two-digit number whose digits sum to 3, that is, it must be 12, 21, or 30. The table below shows all the two-digit numbers whose product is one of these three numbers.

| $\mathbf{1 2}$ | $26,62,34,43$ |
| :--- | :--- |
| $\mathbf{2 1}$ | 37,73 |
| $\mathbf{3 0}$ | 56,65 |

Thus, the numbers 13, 31, 26, 62, 34, 43, 37, 73, 56, and 65 are the only two-digit numbers whose transformed number is 3.

c) $1^{st}$ solution: In the second stage of the game, we have a sum of digits, which is always different from 0; therefore, 0 will never be obtained as the transformed number of a three-digit number in this stage. To obtain 0 as the transformed number of some three-digit number in the first stage, this number must have 0 as the units digit, the tens digit, or both. The three-digit numbers that have 0 in both the units and tens places are 100, 200, ..., 900, for a total of 9. The numbers that have 0 only in the units place are of the form $X Y 0$, where $X$ and $Y$ represent digits from 1 to 9. There are $9 \times 9=81$ numbers of this type, and the same reasoning shows that there are 81 three-digit numbers with 0 only in the tens digit. In total, there are $9+81+81=171$ three-digit numbers whose transformed number is 0.

$2^{nd}$ solution: As in the solution above, we conclude that 0 must appear in the units place, the tens place, or both. The hundreds digit can be any digit from $1$ to $9$. After choosing this digit, the tens and units digits can be chosen in 19 different ways; for example, 100, 101, 102, ..., 109, 110, 120, ..., 190 are the 19 possibilities with 1 in the first position. Therefore, the total sought is $9 \times 19=171$.

$3^{rd}$ solution: As in the solution above, we conclude that 0 must appear in the units place, the tens place, or both. There are 90 numbers with 0 in the units place and 90 with 0 in the tens place, as well as 9 that have 0 in both the tens and units places. In total, there are $90+90-9=171$ three-digit numbers whose transformed number is 0.
Cláudia likes to play with numbers of two or more digits. She chooses one of these numbers, multiplies its digits, and if the product has more than one digit, she adds them. She calls the final result the transformed number of the chosen number. For example, the transformed number of 187 is 11, because $1 \times 8 \times 7=56$ and $5+6=11$; the transformed number of 23 is 6, because $2 \times 3=6$.

a) What is the transformed number of 79?

b) Which are the two-digit numbers whose transformed number is 3?

c) How many three-digit numbers have a transformed number of 0?