3. In the number $2 * 0 * 1 * 6 * 07 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,6,7$ (digits can repeat) so that the resulting 11-digit number is divisible by 75. In how many ways can this be done?
# Answer: 432.

Solution. For a number to be divisible by 75, it is necessary and sufficient that it is divisible by 25 and by 3. To ensure divisibility by 25, we can choose 5 as the last digit (1 way) from the available options.

To ensure divisibility by three, we proceed as follows. We will choose three digits arbitrarily (this can be done in $6 \cdot 6 \cdot 6$ ways), and the fifth digit will be chosen so that the sum of all the digits of the number is divisible by 3. Since among the given digits, there are two digits divisible by 3 (0 and 6), two digits that give a remainder of 1 when divided by 3 (4 and 7), and two digits that give a remainder of 2 when divided by 3 (2 and 5), this selection can be made in two ways.

Applying the rule of product, we get that in total $1 \cdot 6 \cdot 6 \cdot 6 \cdot 2=432$ ways.