## Task 25/66

What is the largest multiple of 11 in which none of the digits occurs more than once?
If all ten digits are to occur exactly once, the number must be ten digits long and have a cross sum of 45. A number is divisible by 11 if and only if the difference between the sum $x$ of the digits in even positions and the sum $y$ of the digits in odd positions is zero or a multiple of 11. Thus, we have the system of equations

$$
x+y=45 \quad ; \quad x-y=11 k
$$

with $k=0 ; \pm 1 ; \pm 2, \ldots$ to be solved in positive integers. This results in

$$
x=\frac{45+11 k}{2} \quad ; \quad y=\frac{45-11 k}{2}
$$

From this, it follows: 1. $k$ is odd; 2. $|k|<5$ (otherwise either $x$ or $y$ would be negative). Therefore, only the values $k= \pm 1$ and $k= \pm 3$ are possible.

I. $k= \pm 1$ yields $x=28, y=17$ or $x=17, y=28$.

II. $k= \pm 3$ yields $x=39, y=6$ or $x=6, y=39$.

Case II does not provide a valid solution, as five different digits always yield a sum greater than 6.

Therefore, (Case I) the ten digits must be divided into two groups of five such that one group has a sum of 28 and the other a sum of 17. The largest number sought will then result if the digits in each group are arranged in descending order (from left to right) and as many odd digits as possible are used in the even-positioned group, and as many even digits as possible are used in the odd-positioned group (where the positions are counted from the units place):

even-positioned group: 97543

odd-positioned group: 86210

sought number: 9876524130