The number of Júlia's house has exactly three digits, whose sum is 24. Find all possible house numbers of Júlia in each of the following situations.

(a) The three digits are the same.

(b) Only two digits are the same.

(c) The digits are all different.
The number 24 must be written as a sum of three digits. Initially, note that the digits $0,1,2,3,4$ and 5 cannot be used. Indeed, if one of them were used, for example the digit 5, then we would need to find two digits whose sum is 19, since $24-5=19$; but we know that this is not possible. The same applies to the digits $0,1,2,3$ and 4. Therefore, the number of Júlia's house can only be composed of the digits $6,7,8$ and 9.

(a) If the three digits are equal, then the number of Júlia's house is 888.

(b) If only two of these digits are equal, these two digits must be equal to 9 and the third must be 6, obtaining the numbers 699, 969 and 996. Indeed, with two digits 8, we fall back to the previous case, and with two 6 or two 7, the sum is at most 14, leaving at least 10 for the third, which is not possible.

(c) If the three digits are distinct, then these digits are 7, 8, and 9. Indeed, if a 6 occurs, the sum of the other two must be $24-6=18$, so we need two 9s and fall back to the previous case. Thus, only the following alternatives remain: $789,798,879,897,978$ and 987.