Divisors leaving remainder 31 for 1186

If a number N leaves a remainder of 31 when dividing 1186, then 1186 - 31 = 1155 must be a perfect, exact multiple of N. Find the total number of divisors for 1155 using prime factorization: 1155 = 5 × 231 = 5 × 3 × 77 = 3^1 × 5^1 × 7^1 × 11^1

Calculate the total number of factors using the exponent formula: Total Factors = (1+1)(1+1)(1+1)(1+1) = 2^4 = 16 factors.

Crucial Constraint Check: The divisor ⟨MATH⟩N⟨/MATH⟩ must be strictly greater than the stated remainder (⟨MATH⟩N > 31⟨/MATH⟩). Filter out all factors of 1155 that are \le 31: Factors to exclude: 1, 3, 5, 7, 11, 15 (3 × 5), 21 (3 × 7), 33... Wait, 33 is > 31. Let's re-verify all small factors systematically:

The complete list of factors \le 31 is: {1, 3, 5, 7, 11, 15, 21}. There are exactly 7 factors to exclude. Valid natural numbers = Total Factors - Excluded Factors = 16 - 7 = 9 numbers.

Answer: (d).