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

1. A. 6
2. B. 7
3. C. 8
4. D. 9

Answer: D

Explanation

Let the natural number (divisor) be ‘d’. According to the division algorithm, Dividend = Divisor × Quotient + Remainder. Here, 1186 = d × Q + 31. This implies d × Q = 1186 – 31 = 1155. A crucial condition for division is that the divisor ‘d’ must be greater than the remainder, so d > 31. We need to find the factors of 1155 that are greater than 31.
First, find the prime factorization of 1155: 1155 = 3 × 5 × 7 × 11.
Now, list all factors of 1155 and identify those greater than 31:
Factors: 1, 3, 5, 7, 11, 15 (3×5), 21 (3×7), 33 (3×11), 35 (5×7), 55 (5×11), 77 (7×11), 105 (3×5×7), 165 (3×5×11), 231 (3×7×11), 385 (5×7×11), 1155 (3×5×7×11).
Factors greater than 31 are: 33, 35, 55, 77, 105, 165, 231, 385, 1155. There are 9 such natural numbers.