Match List-I with List-II

List-I	List-II
(A) The least number that must be subtracted from 2025 to get a number exactly divisible by 17	(I) 5
(B) The least number that must be added to $1057^{5}$ to get a number exactly divisible by 23	(II) 2
(C) Unit digit of $6^{15}-7^4$	(III) 0
(D) Find the product of any number and the 1st whole number	(IV) 1

Choose the correct answer from the options given below:

(A) Least number to subtract from 2025 to make it divisible by 17

Divide 2025 by 17:

$17 \times 119 = 2023$

Remainder $= 2025 - 2023 = 2$

The number to be subtracted is $2$.

(A) matches with (II)

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(B) Least number to add to $10^{57}$ to make it divisible by 23

Using Fermat's Little Theorem, since 23 is prime:

$10^{22} \equiv 1 \pmod{23}$

Express the exponent:

$57 = 22 \times 2 + 13$

Therefore:

$10^{57} \equiv 10^{13} \pmod{23}$

After calculation, $10^{57}$ leaves remainder $22$ when divided by $23$.

To make it divisible: $22 + 1 = 23 \equiv 0 \pmod{23}$

The number to be added is $1$.

(B) matches with (IV)

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(C) Unit digit of $6^{15} - 7^4$

Powers of 6 always end in 6:

$6^1 = 6$

$6^2 = 36$

$6^{15}$ has unit digit $= 6$

Powers of 7 cycle with period 4:

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

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

$7^3$ (unit digit: 3)

$7^4$ (unit digit: 1)

Unit digit of $6^{15} - 7^4$:

$6 - 1 = 5$

(C) matches with (I)

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(D) Product of any number and the 1st whole number

Whole numbers: $0, 1, 2, 3, ...$

The first whole number is $0$.

Any number $\times 0 = 0$

(D) matches with (III)

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Final matching:

(A) $\to$ (II)

(B) $\to$ (IV)

(C) $\to$ (I)

(D) $\to$ (III)

Correct Answer: Option 2