Pat summed all four-digit numbers, each of which contained all the digits $1,2,3$ and 4, and arrived at the sum of 58126.

Mat Pat pointed out that the result was not correct, and at the same time revealed to him that the sum could be obtained in a simpler way than by listing and adding all the numbers. Pat took the advice, solved the problem, and found out that he had originally calculated correctly but had forgotten two addends.

Determine which numbers Pat originally forgot.

(L. Hozová)
Among the 24 four-digit numbers containing all the given digits:

| 1234 | 1243 | 1324 | 1342 | 1423 | 1432 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 2134 | 2143 | 2314 | 2341 | 2413 | 2431 |
| 3124 | 3142 | 3214 | 3241 | 3412 | 3421 |
| 4123 | 4132 | 4213 | 4231 | 4312 | 4321 |

Each of the 4 digits appears exactly 6 times in each place (6 * 4 = 24). The sum of the digits in the units place, as well as in the tens, hundreds, and thousands places, is

$$
6 \cdot (1 + 2 + 3 + 4) = 60.
$$

Therefore, the sum of all the numbers is

$$
60 + 10 \cdot 60 + 100 \cdot 60 + 1000 \cdot 60 = 66660.
$$

Since Pat originally got 58126, the sum of the two missing addends must be

$$
66660 - 58126 = 8534.
$$

Since all the numbers consist of digits less than 5, there is no carry-over when adding any two numbers. The digits in the number 8534 can therefore be obtained as follows:

- $8 = 4 + 4$,
- $5 = 2 + 3$ (the option $1 + 4$ is excluded because then one of the addends would have 4 in two places),
- $3 = 1 + 2$,
- $4 = 1 + 3$ (the option $2 + 2$ is excluded because then one of the addends would have 2 in two places).

The number 8534 can be expressed only as the sum of the numbers 4213 and 4321. These are the numbers that Pat originally forgot.

Scoring suggestion. 3 points for correctly determining the sum of 66660; 1 point for correctly determining the difference of 8534; 2 points for correctly determining the originally missing numbers 4213 and 4321.

Notes. The number of all four-digit numbers containing four different digits is equal to the number of all permutations of a four-element set, which is $4 \cdot 3 \cdot 2 \cdot 1 = 24$.

The overall correct sum can also be derived by grouping suitable addends: for example, the sum of each number $s$ and the number written in reverse is always 5555 (see $1234 + 4321 = 5555$), and there are clearly 12 such pairs; thus, the sum of all the numbers in question is $12 \cdot 5555 = 66660$.