## Task A-1.5.

How many four-digit numbers divisible by 3 do not contain the digits $2, 4, 6$ or 9 in their decimal representation?
## Solution.

We have six digits available: $0,1,3,5,7$ and 8.

The first digit can be chosen in 5 ways (any digit except zero).

1 point

The second and third digits can each be chosen in 6 ways.

1 point

Let the first three chosen digits form a three-digit number $n$. The last digit $y$ for the desired four-digit number $10 n+y$ can always be chosen in exactly two ways. Since out of the six digits we have, two give a remainder of 0, two give a remainder of 1, and two give a remainder of 2 when divided by 3, exactly two of the numbers

$$
10 n+0, \quad 10 n+1, \quad 10 n+3, \quad 10 n+5, \quad 10 n+7, \quad 10 n+8
$$

are divisible by 3, regardless of the number $n$.

Therefore, the total number of such numbers is $5 \cdot 6 \cdot 6 \cdot 2=360$.