CSAT Solved Papers/ 2025/Q46

2025 CSAT — Q46

Quant Number theory 2.5 marks Medium

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

A 0
B 1
C 2 Answer
D 3

Worked rationale

$1^n = 1$ always. The behaviour of $2^n \pmod 4$ has a tiny case-split:

$n = 1$ : $2^1 = 2$ , so $1 + 2 = 3 \equiv 3 \pmod 4$ .
$n \ge 2$ : $2^n$ is divisible by $4$ , so $2^n \equiv 0$ , giving $1 + 0 = 1 \equiv 1 \pmod 4$ .

So as $n$ ranges over all naturals, $(1^n + 2^n) \bmod 4$ takes exactly the values $\{3, 1\}$ — two distinct remainders.

Answer: (c) 2.

Why the other options miss

A

solved the wrong question: thinks the expression is always divisible by $4$ (remainder always $0$ ), missing both actual residues.
B

missed a case: evaluates only $n \ge 2$ (always remainder $1$ ) and forgets the special case $n = 1$ giving remainder $3$ .
D

missed a case: imagines $2^n \bmod 4$ cycles through $\{2, 0, \dots\}$ producing more residues, not realising it settles to $0$ for all $n \ge 2$ .

Specialist insight

The crux is that $2^n \pmod 4$ does not cycle — it is $2$ once ( $n=1$ ) and then $0$ forever ( $n \ge 2$ ). Most modular-power items have a periodic cycle; this one has a transient then a constant, and the entire difficulty is remembering the lone $n=1$ case. Always test the smallest exponents explicitly ( $n=1,2$ ) before declaring a pattern — here $n=1$ is the only source of the second residue. The answer is the size of the residue set $\{1, 3\}$ , which is $2$ . A quick discipline: when a base shares a factor with the modulus ( $2$ and $4$ ), expect the powers to collapse to $0$ , not to cycle.

The trap, in one line

$2^n \bmod 4$ is $2$ only at $n=1$ then $0$ for all $n\ge2$ — so the sum hits just $\{3,1\}$ , two distinct remainders (don't forget $n=1$ ).

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