## Task 11/66

The five numbers of a lottery draw (1 to 90) are sought, about which the following is stated:

a) All digits from 1 to 9 appear exactly once.

b) Only the three middle numbers are even.

c) The smallest number has a common divisor (different from itself) with the largest number.

d) The cross sum of one number is a quarter of the cross sum of the largest number.

e) The cross sums of two other numbers are in the ratio 1:2.
a) Since five numbers are sought, but only nine digits appear, the smallest number must be a single digit.

b) This smallest number is odd.

c) Since it shares a (different from itself) common divisor with the largest number, it is the number Nine. (all other odd single-digit numbers are prime numbers or one). The common divisor is therefore 3. The cross sum of the largest number is thus divisible by 3.

d) Since the cross sum of the largest number must also be divisible by 4 and the cross sum of a two-digit number with different digits is at most 17 (in this case, even at most 15, since the digit 9 cannot appear again), the cross sum of the largest number is 12. Since the largest number is odd, only 75 or 57 are possible.

The cross sum of another number is $\frac{1}{4} \cdot 12=3$. Therefore, only 12 or 21 are possible for it. According to b), it is even, so 12 is the second number.

e) The remaining digits are 3, 4, 6, 8. Two numbers must be formed from these, whose cross sums have a ratio of 1:2. It is easily tried out (6 possibilities) that this is only possible for the two combinations $3 ; 4$ and $6 ; 8$.

According to b), the third number is 34 (43 is odd). Since the largest number is odd according to b), the possibilities 86 for the fourth and 57 for the largest number are excluded.

The five numbers are therefore $9,12,34,68,75$.