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

(a) 6
(b) 7
(c) 8
(d) 9 ✓ UPSC's answer

Why the answer is (d)

• If 1186 divided by a natural number $n$ gives a remainder of 31, then $1186 = n \times q + 31$ for some integer $q$.

• This implies that $n$ must be a divisor of $1186 - 31 = 1155$.

• Additionally, the divisor $n$ must be strictly greater than the remainder, so $n > 31$.

• The prime factorization of 1155 is $3 \times 5 \times 7 \times 11$.

• The total number of divisors of 1155 is $(1+1)(1+1)(1+1)(1+1) = 16$.

• The divisors of 1155 less than or equal to 31 are 1, 3, 5, 7, 11, 15, 21, and 33 (wait, 33 is > 31). Let's list them: 1, 3, 5, 7, 11, 15, 21. (33 is greater than 31). So there are 7 divisors $\le 31$.

• Therefore, the number of divisors greater than 31 is $16 - 7 = 9$.

Why the other options are wrong

(a) 6: Option (a) is incorrect because it undercounts the valid divisors of 1155 that are greater than 31.
(b) 7: Option (b) is incorrect because it fails to account for all 9 divisors of 1155 exceeding the remainder 31.
(c) 8: Option (c) is incorrect because it misses one of the valid divisors of 1155 that is greater than 31.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2023, held on 28 May 2023. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.