UPSC Prelims 2025 CSAT Paper II · Q76 / 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.

Practise this paper free

Practise Prelims PYQs and the daily 21 free

Score the daily set, see your rank, and let Guru explain what you missed.

Start free

Already have an account? Log in