1. Correct Answer: Option D: 9
 2. Explanation:
 To solve this problem, we need to find the natural numbers \( n \) such that when 1186 is divided by \( n \), the remainder is 31. This can be expressed as:
 \[ 1186 \equiv 31 \pmod{n} \]
 This implies:
 \[ 1186 - 31 = 1155 \]
 So, \( n \) must be a divisor of 1155. We need to find all divisors of 1155 that are greater than 31.
 Step-by-step process:
 1. Find the divisors of 1155:
     First, we perform the prime factorization of 1155:
         ○ 1155 is divisible by 3 (since the sum of its digits, 1 + 1 + 5 + 5 = 12, is divisible by 3):
       \[
       1155 \div 3 = 385
       \]
         ○ 385 is divisible by 5 (since it ends in 5):
       \[
       385 \div 5 = 77
       \]
         ○ 77 is divisible by 7 (since 77 divided by 7 equals 11):
       \[
       77 \div 7 = 11
       \]
         ○ 11 is a prime number.
     Therefore, the prime factorization of 1155 is:
     \[
     1155 = 3 \times 5 \times 7 \times 11
     \]
 2. List all divisors of 1155:
     Using the prime factorization, the divisors of 1155 are:
     \[
     1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155
     \]
 3. Select divisors greater than 31:
     From the list of divisors, those greater than 31 are:
     \[
     33, 35, 55, 77, 105, 165, 231, 385, 1155
     \]
     There are 9 such divisors.
 Thus, the number of natural numbers that give a remainder of 31 when 1186 is divided by them is 9.