How many natural numbers are there which give a remainder of 31 when 1186 is divided by these natural numbers?

When 1186 is divided by a natural number 'd', the remainder is 31. According to the division algorithm, we can write: 1186 = d * q + 31 where 'q' is the quotient and 'd' is the divisor. A key condition for division is that the divisor 'd' must be greater than the remainder. So, d > 31.

From the equation, subtract 31 from both sides: 1186 - 31 = d * q 1155 = d * q

This means that 'd' must be a divisor of 1155. We need to find all natural number divisors of 1155 that are greater than 31.

First, find the prime factorization of 1155: 1155 = 3 * 5 * 7 * 11

To find the total number of divisors of 1155, we use the exponents of its prime factors. Since each prime factor (3, 5, 7, 11) has an exponent of 1, the total number of divisors is (1+1)(1+1)(1+1)(1+1) = 2 * 2 * 2 * 2 = 16.

Now, we need to filter these 16 divisors to include only those that are greater than 31. It's easier to list the divisors that are NOT greater than 31 (i.e., less than or equal to 31) and subtract them from the total.

The divisors of 1155 are: 1, 3, 5, 7, 11, 15 (35), 21 (37), 33 (311), 35 (57), 55 (511), 77 (711), 105 (357), 165 (3511), 231 (3711), 385 (5711), 1155 (357*11).

Divisors that are less than or equal to 31 are: 1, 3, 5, 7, 11, 15, 21. There are 7 such divisors.

The number of natural numbers 'd' that satisfy the conditions (d is a divisor of 1155 and d > 31) is: Total number of divisors - Number of divisors <= 31 = 16 - 7 = 9

Thus, there are 9 such natural numbers.

The final answer is D.