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

Question

Accepted Answer

The correct answer is option D. 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.

Q75. How many natural numbers are there which given a remainder of 31 when 1186 is divided by these natural numbers?

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