1. The problem is to find the common divisors of the number 390. 2. First, we find the prime factorization of 390. 3. Divide 390 by the smallest prime number 2: $$390 \div 2 = 195$$, so 2 is a prime factor. 4. Next, divide 195 by 3: $$195 \div 3 = 65$$, so 3 is a prime factor. 5. Next, divide 65 by 5: $$65 \div 5 = 13$$, so 5 is a prime factor. 6. Finally, 13 is a prime number itself. 7. Therefore, the prime factorization of 390 is $$2 \times 3 \times 5 \times 13$$. 8. The common divisors are all the products of these prime factors taken in all possible combinations including 1. 9. The divisors of 390 are: 1, 2, 3, 5, 6 (2*3), 10 (2*5), 13, 15 (3*5), 26 (2*13), 30 (2*3*5), 39 (3*13), 65 (5*13), 78 (2*3*13), 130 (2*5*13), 195 (3*5*13), and 390 (2*3*5*13). 10. These are all the common divisors of 390.