2020 CSAT, all questions

UPSC Prelims 2020 · CSAT

Question 54: Basic Numeracy

What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?

a1012
b1022
c1122
d1222

AnswerUPSC’s answer is (b).

A number leaving remainder 2 on division by 3, 4, 5 and 6 has the form LCM + 2. LCM(3, 4, 5, 6) = 60, so the numbers are 60k + 2. The least four-digit one needs 60k + 2 ≥ 1000, so k = 17 (60 × 16 = 960 is too small), giving 1020 + 2 = 1022.

Why each option is right or wrong
  • (a) 1012 - 2 = 1010 is not a multiple of 60.
  • (b) Right: 1022 - 2 = 1020 = 60 × 17.
  • (c) 1122 - 2 = 1120 is not a multiple of 60.
  • (d) 1222 - 2 = 1220 is not a multiple of 60.

Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.

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Source: UPSC, Civil Services (Preliminary) Examination 2020, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.