UPSC Prelims 2024 CSAT Paper II · Q54 of 79 Basic Numeracy easy

32^5 + 2^27 is divisible by

  1. (a) 3
  2. (b) 7
  3. (c) 10 ✓ UPSC's answer
  4. (d) 11

Why the answer is (c)

• 32^5 = (2^5)^5 = 2^25.

• 2^25 + 2^27 = 2^25(1 + 2^2) = 2^25(1 + 4) = 5 × 2^25.

• The factor 2^25 supplies a factor 2, and the factor 5 supplies a factor 5.

• Since 10 = 2 × 5, the number is divisible by 10, so option (c) follows.

Why the other options are wrong

(a) 3
32^5 + 2^27 = 5 × 2^25 leaves remainder 1 on division by 3, so it is not divisible by 3.
(b) 7
32^5 + 2^27 = 5 × 2^25 leaves remainder 3 on division by 7, so it is not divisible by 7.
(d) 11
32^5 + 2^27 = 5 × 2^25 leaves remainder 6 on division by 11, so it is not divisible by 11.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2024, held on 16 June 2024. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.

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