Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Numbers, Quantification and Numerical Applications

Question:

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.

Options:

(A)-(I), (B)-(III), (C)-(II), (D)-(IV)

(A)-(IV), (B)-(II), (C)-(II), (D)-(III)

(A)-(II), (B)-(I), (C)-(III), (D)-(IV)

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Correct Answer:

(A)-(IV), (B)-(II), (C)-(II), (D)-(III)

Explanation:

The correct answer is Option (2) → (A)-(IV), (B)-(II), (C)-(II), (D)-(III) **

List-I

List-II

(A) Unit's digit of $2^{11}$

(IV) 8

(B) Unit's digits of $11^{132}$

(II) 1

(C) Remainder when 71 × 73 × 75 is divided by 23

(I) 2

(D) Remainder when $7^{30}$ is divided by 5

(III) 4

Compute each value:

(A) Unit digit of $2^{11}$ Cycle of 2: $2,4,8,6$ (length 4) $11 \bmod 4 = 3$ → 3rd term = 8 → (IV)

(B) Unit digit of $11^{132}$ Unit digit of 11 is 1 → $1^{132}=1$ → (II)

(C) Remainder of $71 \times 73 \times 75$ mod 23

$71 \equiv 2$, $73 \equiv 4$, $75 \equiv 6$ (mod 23)

Product = $2 \cdot 4 \cdot 6 = 48$

$48 \bmod 23 = 2$ → (I)

(D) Remainder of $7^{30}$ mod 5 $7 \equiv 2$ mod 5 → $2^{30}$

Cycle of 2 mod 5: $2,4,3,1$ (length 4)

$30 \bmod 4 = 2$ → 2nd term = 4 → (III)