UPSC Prelims 2025 CSAT Paper II · Q76 of 78 Basic Numeracy easy

If n is a natural number, then what is the number of distinct remainders of (1ⁿ + 2ⁿ) when divided by 4 ?

  1. (a) 0
  2. (b) 1
  3. (c) 2 ✓ UPSC's answer
  4. (d) 3

Why the answer is (c)

• 1ⁿ is always congruent to 1 modulo 4 for every natural number n.

• 2¹ is congruent to 2 modulo 4, while 2ⁿ is congruent to 0 modulo 4 for every n ≥ 2.

• For n = 1, (1¹ + 2¹) = 3, giving remainder 3 on division by 4.

• For n ≥ 2, (1ⁿ + 2ⁿ) ≡ 1 + 0 = 1, giving remainder 1 on division by 4.

• The distinct remainders are 1 and 3, so the number of distinct remainders is 2, which is option (c).

Why the other options are wrong

(a) 0
The expression produces remainders 1 and 3, so the count cannot be 0.
(b) 1
The expression gives two different remainders, 3 for n = 1 and 1 for n ≥ 2, so the count is not 1.
(d) 3
Only the remainders 1 and 3 occur, so the count is not 3.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.

Reading the answer is not the same as getting it right under a clock. Practise this question with UPSC's negative marking, and anything you miss goes into an error notebook until you get it right twice.

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