IAS 2023 — CSAT Maths

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 $d$d divides 1186 and leaves a remainder of 31, then it must satisfy the condition:

$1186 = q \times d + 31$1186=q×d+31

where $q$q is the quotient and $d > 31$d>31. The condition $d > 31$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$1186−31=q×d

$1155 = q \times d$1155=q×d

This implies that $d$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$1155=3×5×7×11

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