CSAT Solved Papers/ 2023/Q74

2023 CSAT — Q74

Quant Number theory 2.5 marks Easy

What is the remainder if $2^{192}$ is divided by $6$ ?

A 0
B 1
C 2
D 4 Answer

Worked rationale

List the powers of $2$ mod $6$ to find the cycle:

2^1 \equiv 2,\quad 2^2 \equiv 4,\quad 2^3 \equiv 8 \equiv 2,\quad 2^4 \equiv 4,\ \dots \pmod 6.

From $2^1$ onward the residues alternate $2, 4, 2, 4, \dots$ — odd exponents give $2$ , even exponents give $4$ .

Since $192$ is even,

2^{192} \equiv 4 \pmod 6.

Answer: (d) 4.

Why the other options miss

A

solved the wrong question: assumes $6 \mid 2^{192}$ because $2 \mid 2^{192}$ , forgetting $6 = 2\cdot 3$ and $3 \nmid 2^{192}$ .
B

a remainder-rule slip: misapplies a Fermat/Euler idea ( $2^{\phi(6)} \equiv 1$ is false since $\gcd(2,6)\ne 1$ ), wrongly concluding remainder $1$ .
C

off by one: reads the cycle for an odd exponent (giving $2$ ), mismatching the even $192$ .

Specialist insight

The trap is that Euler’s theorem does not apply — $2$ and $6$ are not coprime — so don’t reach for $2^{\phi(6)}\equiv 1$ . Instead just list the residues: $2^n \bmod 6$ is $2$ for odd $n$ and $4$ for even $n$ (for $n \ge 1$ ). The exponent $192$ is even, so the answer is $4$ . A two-line cycle table beats any theorem here.

The trap, in one line

$2^n \bmod 6$ is $2$ (odd $n$ ) or $4$ (even $n$ ); $192$ even $\Rightarrow 4$ — Euler's theorem doesn't apply ( $\gcd(2,6)\ne 1$ ) $=$ (d).

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