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

EXPLANATION

Correct Option (C)

To determine the natural numbers that yield a remainder of 31 when 1186 is divided by them, we apply the division algorithm. If a natural number $d$ divides 1186 and leaves a remainder of 31, then it must satisfy the condition:

$1186 = q \times d + 31$

where $q$ is the quotient and $d > 31$ . The condition $d > 31$ is essential because the remainder must always be strictly less than the divisor.

Rearranging the equation, we get:

$1186 - 31 = q \times d$

$1155 = q \times d$

This implies that $d$ must be a divisor of 1155. We need to identify all positive divisors of 1155 that are greater than 31.

First, we find the prime factorization of 1155:

$1155 = 3 \times 5 \times 7 \times 11$

The divisors of 1155 are formed by combining these prime factors. We then select only those divisors that are greater than 31:

$3 \times 11 = 33$
$5 \times 7 = 35$
$5 \times 11 = 55$
$7 \times 11 = 77$
$3 \times 5 \times 7 = 105$
$3 \times 7 \times 11 = 231$
$5 \times 7 \times 11 = 385$
$3 \times 5 \times 7 \times 11 = 1155$

Counting these numbers, we find there are 8 such natural numbers that satisfy the given conditions.

Incorrect Options:

Options (A) 6, (B) 7, and (D) 9 are incorrect because a systematic calculation of the divisors of 1155 that are strictly greater than 31 yields exactly 8 numbers. Any other count would result from an incomplete or incorrect identification of the valid divisors or a misapplication of the remainder condition.