What is the remainder if 2192 is divided by 6?

Correct Option (D)

To determine the remainder when $2^{192}$ is divided by 6, we can observe the pattern of remainders for successive powers of 2:

• $2^1÷6$: Remainder = 2
• $2^2÷6$: $4÷6$: Remainder = 4
• $2^3÷6$: $8÷6$: Remainder = 2
• $2^4÷6$: $16÷6$: Remainder = 4

A clear pattern emerges for the remainders: 2, 4, 2, 4, ... This cycle of remainders has a length of 2.

For any integer exponent \(n \ge 1\):

• If $n$ is odd, the remainder is 2.
• If $n$ is even, the remainder is 4.

Since the exponent 192 is an even number, the remainder when $2^{192}$ is divided by 6 is 4.

Incorrect Options:

The established pattern of remainders for $2^n÷6$ (for \(n \ge 1\)) is consistently 2 or 4.

• Option (A) 0: A remainder of 0 would imply that $2^{192}$ is perfectly divisible by 6. This is not possible because $2^{192}$ is a power of 2 and thus contains only prime factors of 2, whereas 6 requires a prime factor of 3 for divisibility.
• Option (B) 1: The remainder 1 does not occur in the observed cycle of remainders for powers of 2 divided by 6.
• Option (C) 2: The remainder 2 occurs when the exponent is odd. Since 192 is an even exponent, 2 is not the correct remainder in this instance.