## Task 2 - 280712

From the digits 1, 3, 4, 5, 7, and 9, six-digit natural numbers are to be formed, in which each of these digits appears exactly once.

a) Determine the number of all different numbers that can be formed in this way.

b) Investigate which of the numbers formed in this way are divisible by 18!
a) With two digits, exactly two different two-digit numbers can be formed (e.g., 13 and 31). (1) If a third digit is added, it can be placed before, between, or after the two digits of the two-digit number; this results in three possibilities. Therefore, from (1), because $2 \cdot 3=6$, exactly six different three-digit numbers can be formed from three different digits. (2)

If four-digit numbers are formed by adding a fourth digit, this fourth digit can stand in the first, second, third, or fourth position (four possibilities), and from (2), because $6 \cdot 4=24$, 24 four-digit numbers can be formed in this way. (3)

By analogous reasoning, it is recognized that there are exactly five possibilities for the position of a fifth digit in a four-digit number, and thus, because of (3) and $5 \cdot 24=120$, a total of 120 five-digit numbers can be formed. (4)

The number of six-digit numbers follows analogously from (4) and because $6 \cdot 120=720$. Therefore, exactly 720 different six-digit numbers can be formed from the given digits.
b) A natural number is only divisible by 18 if it is divisible by 9. This is only the case if its cross sum is divisible by 9.

However, all 720 numbers formed in the required manner have the cross sum $1+3+4+5+7+9=29$, so none of these numbers is divisible by 18.