## Task 8/83

How many natural numbers divisible by 4 can be formed from the 9 digits $1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9$ if each of these digits must appear exactly once in each number (assuming decimal notation)?
A natural number $n$ is (as is well known) divisible by 4 without a remainder if and only if the number formed by its last two digits is. From the given digits, 16 two-digit numbers can be formed that are divisible by 4:

In front of each of these 16 numbers, the remaining 7 digits can be arranged in 7! different ways. Consequently, for the number $k$ of possible numbers, we have $k=7! \cdot 16 = 8! \cdot 2 = 80640$.