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 10-digit hussel numbers are there?
## 1. Version for Grade 5 & Grade 6

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, it must be 2, 4, 6, or 8 because 0 is not included due to property (1). Since we can place any 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 rearranging 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 its digits is also divisible by 3. If we have \( k \) eights and \( 10 - k \) fours, then the sum of the digits is \( 8k + 4(10 - k) = 40 + 4k = 36 + 4(k + 1) \). This is divisible by 3 if \( k + 1 \) is divisible by 3, i.e., if \( k = 2 \), \( k = 5 \), or \( k = 8 \). We see that a hussel number of ten digits must consist of 2 eights and 8 fours, 5 eights and 5 fours, or 8 eights and 2 fours. Note that these numbers all satisfy properties (1) and (3). We only need to check which of these numbers are divisible by 11 (property (2)).

We use the 11-criterion: a number is divisible by 11 if the alternating sum of its digits is divisible by 11. By alternating sum, we mean that we take the digits alternately with a plus sign and a minus sign. In our case, all digits are either 4 or 8, so the alternating sum must be divisible by \( 4 \times 11 = 44 \). Since the alternating sum can never be greater than \( 8 + 8 + 8 + 8 + 8 - 4 - 4 - 4 - 4 - 4 = 20 \) (and never less than -20), the alternating sum must be 0. In other words, the sum of the five digits in the odd positions must be equal to the sum of the five digits in the even positions. This means that the number of eights in the odd positions must be equal to the number of eights in the even positions. We consider the three cases we previously distinguished:

- Suppose there are 2 eights in total. The only requirement now is that there is exactly 1 eight in an odd position and 1 eight in an even position. There are thus \( 5 \times 5 = 25 \) possibilities.
- Suppose there are 5 eights in total. Since we cannot divide the eights into two equal quantities, there are no solutions in this case.
- Suppose there are 8 eights in total. The only requirement now is that there are exactly 4 eights in the odd positions and 4 eights in the even positions. In other words, 1 odd position is a four and 1 even position is a four. There are again \( 5 \times 5 = 25 \) possibilities.

We conclude that there are a total of \( 25 + 25 = 50 \) ten-digit hussel numbers.