Sofia plays by writing all the four-digit numbers with different digits that can be written with the digits 1, 2, 4, and 7. She adds three of these numbers - all different - and gets 13983. What are these three numbers?
Since 13983 ends in 3, the sum of the units digits of the three different numbers must be 13 or 23. Since 23 cannot be obtained by adding 1, 2, 4, and 7, we only have one option, namely, $2+4+7=13$.

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-092.jpg?height=291&width=331&top_left_y=374&top_left_x=1522)

Now, the sum of the tens digits must be $8-1=7$ and, therefore, can only be $1+2+4=7$. We complete the tens digits, being careful not to repeat the same digit in the same number. We have only the following three options.
![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-092.jpg?height=290&width=1182&top_left_y=924&top_left_x=518)

The hundreds digits must sum to 9, which leaves us with two possibilities, $4+4+1$ or $1+1+7$. Since in all three options the digit 4 appears in two of the three numbers, we choose the possibility $1+1+7$ for the hundreds, so that the digit 4 does not appear repeated. We also need to ensure that the digits 1 and 7 do not appear repeated, which eliminates the third option above and leads us to two options for the hundreds, as follows.
![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-092.jpg?height=304&width=936&top_left_y=1702&top_left_x=638)

Finally, the thousands digits must sum to 13, and it is easy to choose them. Thus, Sofia can reach 13983 in two ways, as follows.

|  | 4 | 7 | 1 | 2 |
| :---: | :---: | :---: | :---: | :---: |
|  | 7 | 1 | 2 | 4 |
|  | 2 | 1 | 4 | 7 |
| 1 | 3 | 9 | 8 | 3 |


|  | 7 | 1 | 4 | 2 |
| :---: | :---: | :---: | :---: | :---: |
|  | 2 | 7 | 1 | 4 |
|  | 4 | 1 | 2 | 7 |
| 1 | 3 | 9 | 8 | 3 |