Match List-I with List-II

Choose the correct answer from the options given below:

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

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

(A): Unit's digit of $(257)^{153} \times (346)^{72}$

• Unit's digit of $(257)^{153}$: This depends only on the last digit, $7$. The unit's digits of powers of $7$ follow a cycle of four: $7, 9, 3, 1$. Since $153 \div 4$ leaves a remainder of $1$, the unit's digit is $7^1 = \mathbf{7}$.
• Unit's digit of $(346)^{72}$: Any power of a number ending in $6$ will always end in $6$.
• Product: $7 \times 6 = 42$. The unit's digit is $2$.
• Match: (A) $\rightarrow$ (IV)

(B): Unit's place of the product $61 \times 62 \times \dots \times 69$

• In this sequence, we have $62$ (which ends in $2$) and $65$ (which ends in $5$).
• Multiplying $2 \times 5 = 10$. Any product that includes a multiple of $10$ will have $0$ as its unit's digit.
• Match: (B) $\rightarrow$ (III)

(C): Remainder when $121012$ is divided by $12$

• $121012 = 120000 + 1012$.
• Since $120000$ is perfectly divisible by $12$, we only need to check $1012 \div 12$.
• $1012 = (12 \times 80) + 52$.
• $52 = (12 \times 4) + 4$.
• The remainder is $4$.
• Match: (C) $\rightarrow$ (II)

(D): Unit's digit of $(257)^{153} + (346)^{72}$

• From Item (A), we know the unit's digit of $(257)^{153}$ is $7$.
• The unit's digit of $(346)^{72}$ is $6$.
• Sum: $7 + 6 = 13$. The unit's digit is $3$.
• Match: (D) $\rightarrow$ (I)