4. If you add the digits of the number 2022, you get 6.

How many 4-digit numbers (including 2022 itself) are there such that when you add the digits, you get 6? The numbers may not start with a 0.
A) 40
B) 45
C) 50
D) 55
E) 56
A4. E) 56 For the four digits of the number, there are not so many possibilities. We will go through them all.

Once 6 and three times 0. A four-digit number cannot start with 0, so there is only 1 possibility here.

Once 5, once 1, and twice 0. We can start with the 1, and then there are 3 possibilities for the position of the 5. Conversely, we can also start with the 5, and then there are 3 possibilities for the position of the 1. Together, there are thus $3+3=6$ ways.

Once 4, once 2, and twice 0. With exactly the same reasoning as in the previous case, there are 6 possibilities.

Once 4, twice 1, and once 0. We choose a position for the 0 that is not at the beginning; this can be done in 3 ways. For the 4, there are then 3 remaining possibilities, and the other two digits are 1s. In total, there are thus $3 \times 3=9$ possibilities.

Twice 3 and twice 0. We must start with a 3. For the other 3, there are three remaining positions; this can thus be done in 3 ways.

Once 3, once 2, once 1, and once 0. There are 3 possibilities for the 0, then 3 possibilities for the 1, 2 for the 2, and the remaining digit must be a 3. This results in $3 \times 3 \times 2=18$ possibilities.

Once 3 and three times 1. The 3 can be in any of the four possible positions, and the remaining digits are then 1s. This can be done in 4 ways.

Three times 2 and once 0. The 0 must come in one of the last three positions; this can be done in 3 ways.

Twice 2 and twice 1. For the first 1, we can choose 4 possible positions, and for the second 1, there are still 3. But note: we have counted each possibility for both 1s twice (since it doesn't matter which 1 is the first). There are thus $\frac{1}{2} \times 4 \times 3=6$ possibilities.

In total, we have found $1+6+6+9+3+18+4+3+6=56$ solutions.