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UPSC CSE CSAT 2023 · Paper II (CSAT) · Question 45 · Number system

ऐसे कितने धन पूर्णांक हैं जिनसे 1186 को विभाजित करने पर शेषफल 31 आता है ?

UPSC Civil Services (Preliminary) Examination 2023, General Studies Paper II (CSAT), Question 45 — Number system.

CSAT 2023 · Q45

Number system X-Ray · Core #4 Medium

ऐसे कितने धन पूर्णांक हैं जिनसे 1186 को विभाजित करने पर शेषफल 31 आता है ?

Read this question in English

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

उत्तर और व्याख्या · explanation in English

Answer: (d) 9

If 1186 leaves remainder 31, the divisor divides 1186 − 31 = 1155 and must be greater than 31. 1155 = 3 × 5 × 7 × 11 has 16 divisors, of which 9 are greater than 31.

  1. 1186 = n × (quotient) + 31, so n divides 1186 − 31 = 1155; also n > 31, because a remainder is always less than the divisor.
  2. 1155 = 3 × 5 × 7 × 11, so it has 2 × 2 × 2 × 2 = 16 divisors.
  3. Divisors greater than 31: 33, 35, 55, 77, 105, 165, 231, 385, 1155 — 9 numbers.

Remember · Remainder r from N ÷ n means n divides N − r and n > r; count divisors of N − r above r.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2023, Q45 — minimalistias.com/pyqs/csat/2023/q45

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