
1. We call a positive integer a shuffle number if the following hold:

(1) All digits are nonzero.

(2) The number is divisible by 11 .

(3) The number is divisible by 12. If you put the digits in any other order, you again have a number that is divisible by 12 .

How many 10-digit shuffle numbers are there?

## 1. Version for klas $4 \&$ below

First observe that a shuffle number can only contain the digits 2, 4, 6, and 8 . Indeed, if we place the digits in any order, we obtain an even number (since it is divisible by 12) because of property (3). Since the last digit of an even number is also even, the digit must be $2,4,6$, or 8 since it cannot be 0 because of property (1). Since we can put any of the digits in the last position, this holds for each digit of a shuffle number.

Next, observe that a shuffle number can only contain the digits 4 and 8 . Indeed, suppose that we had a shuffle number containing the digit 2 . We could reorder the digits so that the last digit is 2 . The last two digits would then be $22,42,62$, or 82 . But then the number would not be divisible by 4 (and hence also not divisible by 12), contradicting property (3). In the same way we see that a shuffle number cannot contain digit 6 because a number ending in digits $26,46,66$, or 86 is not divisible by 4 .

A shuffle number is divisible by 3 (since it is divisible by 12), hence the sum of its digits is divisible by 3 as well. 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: $\mathbf{3 6}$
- 0 fours and 5 eights. Sum of the digits: 40

The 5-digit numbers for which properties (1) and (3) hold, are precisely the numbers that have 4 fours and 1 eight, or have 1 four and 4 eights. It remains to examine which of these eight numbers are divisible by 11. For this we use 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 instead of adding them, we alternately add and subtract. For the eight candidate solutions 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 | $8-8+8-8+4=4$ |

We see that there are precisely two 5-digit shuffle numbers, namely 48444 and 44484.

