1. Natural numbers, for which the sum of the digits equals 5, were arranged in ascending order. What number is in the $122-$nd position?
Answer: 40001.

Solution: Let's calculate the number of such numbers for different digit counts.

Let $n$ be the number of digits. Subtract 1 from the most significant digit, we get a number (which can now start with zero), the sum of whose digits is 4. Represent this as follows - there are 4 balls, between which $n-1$ partitions are placed. This can be chosen in $C_{n+3}^{4}$ ways.

Let's construct a table:

| Number of digits | Number of ways |
| :--- | :---: |
| 1 | $C_{4}^{4}=1$ |
| 2 | $C_{5}^{4}=5$ |
| 3 | $C_{6}^{4}=15$ |
| 4 | $C_{7}^{4}=35$ |
| 5 | $C_{8}^{4}=70$ |
| TOTAL | $\mathbf{1 2 6}$ |

Obviously, the 126th number - the largest 5-digit number - will be 50000.

Thus, the 125th number is 41000.

the 124th is 40100

the 123rd is 40010

the 122nd is 40001