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

Correct Answer: (D) 9

Dividing 1186 by natural number $n$ leaves remainder 31, which means $n$ must divide $1186 - 31 = 1155$, with the constraint that $n > 31$. Prime factorizing 1155 gives $3$$\times$ 5 $\times$ 7 $\times$ 11, which has $(1+1)(1+1)(1+1)(1+1) = 16$ total divisors. Listing the divisors less than or equal to 31: 1, 3, 5, 7, 11, 15, and 21 (7 divisors). Subtracting these from the total gives $16 - 7 = 9$ natural numbers strictly greater than 31 (namely 33, 35, 55, 77, 105, 165, 231, 385, and 1155).