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 × 73 × 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.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(II), (C)-(II), (D)-(III)
(A) Unit's digit of \(2^{11}\)
This matches (IV) 8. The unit digits of powers of 2 repeat in the cycle 2, 4, 8, 6 (length 4). Since \(11 = 4 \times 2 + 3\), the unit digit is the third in the cycle:
\(2^{11} = 2048\), unit digit 8
(B) Unit's digits of \(11^{132}\)
This matches (II) 1. Every power of a number ending in 1 ends in 1.
(C) Remainder when 71 × 73 × 75 is divided by 23
This matches (I) 2. Since \(69 = 3 \times 23\), reduce each factor modulo 23:
\(71 \equiv 2,\quad 73 \equiv 4,\quad 75 \equiv 6 \pmod{23}\)
\(71 \times 73 \times 75 \equiv 2 \times 4 \times 6 = 48 \pmod{23}\)
\(48 = 2 \times 23 + 2\), remainder 2
(D) Remainder when \(7^{30}\) is divided by 5
This matches (III) 4.
\(7 \equiv 2 \pmod 5\), and \(2^4 = 16 \equiv 1 \pmod 5\)
\(7^{30} \equiv 2^{30} = (2^4)^7 \times 2^2 \equiv 1 \times 4 = 4 \pmod 5\)
So the matching is (A)-(IV), (B)-(II), (C)-(I), (D)-(III).
Note: It seems that there is a mistake in the original NTA paper, which prints (C)-(II) in Option (2) when it should read (C)-(I): the remainder is 2, not 1. The question can still be solved because (A)-(IV), (B)-(II) and (D)-(III) are correct, and Option (2) is the only option containing them.