1. Call a positive integer a hussel number if:

(1) All digits are not equal to 0.

(2) The number is divisible by 11.

(3) The number is divisible by 12. If you rearrange the digits in any other random order, you always get a number that is divisible by 12.

How many 5-digit hussel numbers are there?
## 1. Version for Class 4 and Lower

To begin with, we note that a hussel number can only contain the digits 2, 4, 6, and 8. Indeed, if we arrange the digits in any random order, we get an even number (divisible by 12) due to property (3). Since the last digit of an even number is always even, that digit must be 2, 4, 6, or 8 because 0 is not included due to property (1). Since we can place each digit at the end, this applies to every digit of a hussel number.

Next, we note that a hussel number can only contain the digits 4 and 8. Suppose a hussel number has the digit 2. By suitably swapping the digits, we get a number where the last digit is 2. The last two digits of this number are then 22, 42, 62, or 82. The number is then not divisible by 4 (and thus not divisible by 12), contradicting property (3). Similarly, a hussel number cannot have the digit 6 because a number ending in 26, 46, 66, or 86 is not divisible by 4.

Since a hussel number is divisible by 3 (because it is divisible by 12), the sum of the digits is also divisible by 3. Since each digit is either 4 or 8, we have the following cases:

- 5 fours and 0 eights. Sum of the digits: 20
- 4 fours and 1 eight. Sum of the digits: 24
- 3 fours and 2 eights. Sum of the digits: 28
- 2 fours and 3 eights. Sum of the digits: 32
- 1 four and 4 eights. Sum of the digits: 36
- 0 fours and 5 eights. Sum of the digits: 40

The five-digit numbers that satisfy properties (1) and (3) are therefore exactly the numbers that consist of 4 fours and 1 eight, or 1 four and 4 eights. We now need to check which of these eight numbers are divisible by 11. This is simple using the 11-criterion: a number is divisible by 11 if the alternating sum of the digits is divisible by 11. By alternating sum, we mean that we take the digits alternately with a plus sign and a minus sign. For the eight candidates, we find the following alternating sums:

| 84444 | $8-4+4-4+4=8$ | 48888 | $4-8+8-8+8=4$ |
| ---: | :--- | :--- | :--- |
| $\mathbf{4 8 4 4 4}$ | $4-8+4-4+4=0$ | 84888 | $8-4+8-8+8=12$ |
| 44844 | $4-4+8-4+4=8$ | 88488 | $8-8+4-8+8=4$ |
| $\mathbf{4 4 4 8 4}$ | $4-4+4-8+4=0$ | 88848 | $8-8+8-4+8=12$ |
| 44448 | $4-4+4-4+8=8$ | 88884 | $8-8+8-8+4=4$ |

Thus, there are exactly two hussel numbers of 5 digits, namely 48444 and 44484.