Match List-I with List-II

List-I	List-II
(A) Unit's digit of $(257)^{153} \times (346)^{72}$	(I) 3
(B) Unit's place of the product $61 \times 62 \times 63 \times 64 \times ... \times 69$ is	(II) 4
(C) 121012 is divided by 12, the remainder is	(III) 0
(D) Unit's digit of $(257)^{153} + (346)^{72}$	(IV) 2

Choose the correct answer from the options given below:

Finding the unit's digit of $(257)^{153} \times (346)^{72}$:

For multiplication, find the unit's digit of each number separately, then multiply them.

Unit's digit of 257 is 7. Finding the pattern when multiplying 7 repeatedly:

$7^1 = 7$ (unit's digit: 7)

$7^2 = 49$ (unit's digit: 9)

$7^3 = 343$ (unit's digit: 3)

$7^4 = 2401$ (unit's digit: 1)

$7^5 = 16807$ (unit's digit: 7)

Pattern: {7, 9, 3, 1} repeats every 4 powers.

Dividing the exponent: $153 \div 4 = 38$ remainder $1$

Remainder 1 corresponds to position 1 in pattern: 7

Unit's digit of 346 is 6. Any power of 6 always ends in 6.

Multiplying the unit's digits:

$7 \times 6 = 42$

Unit's digit = 2

(A) → (IV)

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Finding the unit's place of $61 \times 62 \times 63 \times 64 \times ... \times 69$:

To get unit's digit 0, the product needs both 2 and 5 as factors.

$65 = 5 \times 13$ (provides factor 5)

$64 = 2^6$ (provides factors of 2)

Since both are in the product: $2 \times 5 = 10$

Any number with 10 as a factor has unit's digit 0.

(B) → (III)

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Finding the remainder when 121012 is divided by 12:

Breaking down 121012:

$12 \times 10000 = 120000$

$121012 - 120000 = 1012$

Dividing 1012 by 12:

$12 \times 84 = 1008$

$1012 - 1008 = 4$

Remainder = 4

(C) → (II)

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Finding the unit's digit of $(257)^{153} + (346)^{72}$:

From the earlier calculations:

Unit's digit of $(257)^{153} = 7$

Unit's digit of $(346)^{72} = 6$

Adding the unit's digits:

$7 + 6 = 13$

Unit's digit = 3

(D) → (I)

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Final Matching:

(A) → (IV) = 2

(B) → (III) = 0

(C) → (II) = 4

(D) → (I) = 3