3. 3.1. Find the number of four-digit numbers composed of the digits $1,2,3,4,5,6,7$ (each digit is used no more than once) that are divisible by 15.
Answer: 36.

Solution.

For a number to be divisible by 15, it must be divisible by 3 and by 5. Therefore, according to the divisibility rule for 5, the last digit can only be 5. The sum of the digits of the number must be divisible by 3. For this, the sum of the first three digits must give a remainder of 1 when divided by 3 (since 5 gives a remainder of 2 when divided by 3). Let's exclude 5 and replace the numbers with their remainders when divided by 3: 1, 2, 0, 1, 0, 1. To get a remainder of 1, we can:

1) choose two ones and a two (3 ways);
2) choose one and two zeros (3 ways).

In total, there are 6 ways. Note that we did not count any of the ways twice. The 3 chosen digits can be arranged in the number in $3 \cdot 2 \cdot 1=6$ ways. Therefore, the total number of such numbers is $6 \cdot 6=36$.