IAS 2023 — CSAT Maths

2023mcqmedium

What is the remainder if 2192 is divided by 6?

Official previous-year question

Verified 2 Aug 2026.

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Solution

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:

$2^1 \div 6$ : Remainder = 2
$2^2 \div 6$ : $4 \div 6$ : Remainder = 4
$2^3 \div 6$ : $8 \div 6$ : Remainder = 2
$2^4 \div 6$ : $16 \div 6$ : 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 $2^n \div 6$ (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.

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Solution

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:

$2^1 \div 6$ : Remainder = 2
$2^2 \div 6$ : $4 \div 6$ : Remainder = 4
$2^3 \div 6$ : $8 \div 6$ : Remainder = 2
$2^4 \div 6$ : $16 \div 6$ : 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 $2^n \div 6$ (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.

IAS 2023 — CSAT Maths

Options

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Solution

Correct Option (D)

Incorrect Options:

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IAS 2023 — CSAT Maths

Options

Did you get this right?

Solution

Correct Option (D)

Incorrect Options:

Your note

IAS Maths in other years

More IAS Maths questions

Other CSAT topics for IAS

Comprehension

Reasoning

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