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) The product of any number and the 1st whole number is:	(IV) 1

Choose the correct answer from the options given below:

To find the number that must be subtracted from $2025$ to make it divisible by $17$, we need to find the remainder when $2025$ is divided by $17$.

Divide $2025$ by $17$:

$2025 \div 17 = 119$ with some remainder

Calculate: $17 \times 119 = 2023$

The remainder is: $2025 - 2023 = 2$

Therefore, (A) matches with (II) $= 2$

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To find the number that must be added to $1057^5$ to make it divisible by $23$, we work with remainders.

First, find the remainder when $1057$ is divided by $23$:

$1057 \div 23 = 45$ remainder $22$

So $1057 \equiv 22 \pmod{23}$

Note that $22 = 23 - 1$, so $22 \equiv -1 \pmod{23}$

Therefore:

$1057^5 \equiv (-1)^5 = -1 \equiv 22 \pmod{23}$

To make it divisible by $23$, we need to add:

$23 - 22 = 1$

Therefore, (B) matches with (IV) $= 1$

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To find the unit digit of $6^{15} - 7^4$, we only need to track the last digits.

For $6^{15}$:

Powers of $6$ always end in $6$: ($6^1 = 6$, $6^2 = 36$, $6^3 = 216$...)

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

For $7^4$:

Powers of $7$ cycle through unit digits: $7, 9, 3, 1$ (repeats every $4$ powers)

$7^4$ has unit digit $1$

The unit digit of the subtraction:

$6 - 1 = 5$

Therefore, (C) matches with (I) $= 5$

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Whole numbers are: $0, 1, 2, 3, 4...$

The first whole number is $0$

The product of any number $n$ with $0$ is:

$n \times 0 = 0$

Therefore, (D) matches with (III) $= 0$

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Final matching:

(A) → (II) $= 2$

(B) → (IV) $= 1$

(C) → (I) $= 5$

(D) → (III) $= 0$