When 1186 is divided by a natural number, say 'n', it yields a remainder of 31. This can be expressed as 1186 = qn + 31, where 'q' is the quotient. Rearranging this, we get qn = 1186 - 31, which simplifies to qn = 1155. This means 'n' must be a divisor of 1155. Additionally, the remainder (31) must be less than the divisor ('n'). Therefore, we need to find the number of divisors of 1155 that are greater than 31. The prime factorization of 1155 is 3 x 5 x 7 x 11. The total number of divisors is (1+1)(1+1)(1+1)(1+1) = 16. We list the divisors: 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155. Out of these 16 divisors, those that are greater than 31 are 33, 35, 55, 77, 105, 165, 231, 385, and 1155, which amounts to 9 numbers. Thus, there are 9 such natural numbers.