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UPSC CSE CSAT 2022 · Paper II (CSAT) · Question 65 · Number system

1000 से बड़ी वह लघुतम संख्या कौन-सी है जिसे 6, 9, 12, 15, 18 में से किसी एक से भी विभाजित करें, तो…

UPSC Civil Services (Preliminary) Examination 2022, General Studies Paper II (CSAT), Question 65 — Number system.

CSAT 2022 · Q65

Number system Easy

1000 से बड़ी वह लघुतम संख्या कौन-सी है जिसे 6, 9, 12, 15, 18 में से किसी एक से भी विभाजित करें, तो शेषफल 3 बचे?

Read this question in English

What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?

(a) 1063 · (b) 1073 · (c) 1083 · (d) 1183

उत्तर और व्याख्या · explanation in English

Answer: (c) 1083

A number leaving remainder 3 with each divisor is 3 more than a common multiple of them. LCM(6, 9, 12, 15, 18) = 180, and the first multiple of 180 that gives a number above 1000 is 1080, so the answer is 1083.

  1. LCM(6, 9, 12, 15, 18) = 180.
  2. The number must be of the form 180k + 3.
  3. 180 × 5 + 3 = 903 is below 1000; 180 × 6 + 3 = 1083 is the first above it.
  4. Check: 1083 − 3 = 1080 is divisible by 6, 9, 12, 15 and 18.

Remember · Same remainder r for several divisors: the number = k × LCM + r.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2022, Q65 — minimalistias.com/pyqs/csat/2022/q65

In the UPSC syllabus: Number system › Divisibility, remainders, HCF & LCM See it and tick it covered →

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