Match List-I with List-II

List-I	List-II
(A) Unit's digit of $2^{11}$	(I) 2
(B) Unit's digits of $11^{132}$	(II) 1
(C) Remainder when $71 \times 73 \times 75$ is divided by 23	(III) 4
(D) Remainder when $7^{30}$ is divided by 5	(IV) 8

Choose the correct answer from the options given below:

Powers of 2 follow a repeating pattern for unit digits:

$2^1 = 2$ (unit digit: 2)

$2^2 = 4$ (unit digit: 4)

$2^3 = 8$ (unit digit: 8)

$2^4 = 16$ (unit digit: 6)

$2^5 = 32$ (unit digit: 2)

The pattern {2, 4, 8, 6} repeats every 4 powers.

To find the unit digit of $2^{11}$:

$11 \div 4 = 2$ remainder $3$

The remainder indicates position 3 in the pattern, which is 8.

(A) matches with (IV) = 8

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Any power of 11 always ends in 1:

$11^1 = 11$ (ends in 1)

$11^2 = 121$ (ends in 1)

$11^3 = 1331$ (ends in 1)

This occurs because $11 \times 1 = ...1$ always.

Therefore, $11^{132}$ has unit digit 1.

(B) matches with (II) = 1

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Using modular arithmetic, find what each number leaves when divided by 23:

$71 \div 23 = 3$ remainder $2$, so $71 \equiv 2 \pmod{23}$

$73 \div 23 = 3$ remainder $4$, so $73 \equiv 4 \pmod{23}$

$75 \div 23 = 3$ remainder $6$, so $75 \equiv 6 \pmod{23}$

Multiply the remainders:

$2 \times 4 \times 6 = 48$

Divide 48 by 23:

$48 \div 23 = 2$ remainder $2$

(C) matches with (I) = 2

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Find the pattern of powers of 7 modulo 5:

$7^1 = 7$, remainder when divided by 5 is $2$

$7^2 = 49$, remainder when divided by 5 is $4$

$7^3 = 343$, remainder when divided by 5 is $3$

$7^4 = 2401$, remainder when divided by 5 is $1$

$7^5$ has remainder $2$ (pattern repeats)

The pattern {2, 4, 3, 1} repeats every 4 powers.

To find remainder for $7^{30}$:

$30 \div 4 = 7$ remainder $2$

Position 2 in the pattern corresponds to 4.

(D) matches with (III) = 4

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Final matching:

(A) → (IV) 8

(B) → (II) 1

(C) → (I) 2

(D) → (III) 4