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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Each List-I item is worked out on its own: the first two with remainders, the third with units digits, and the fourth from the meaning of whole numbers.
(A) The least number that must be subtracted from 2025 to get a number exactly divisible by 17 — this matches (II) 2.
\(17\times119=2023\), so \(2025=17\times119+2\)
Taking away the remainder 2 leaves 2023, a multiple of 17
(B) The least number that must be added to \(1057^5\) to get a number exactly divisible by 23 — this matches (IV) 1.
\(23\times46=1058\), so 1057 is 1 less than a multiple of 23: \(1057=1058-1\)
In \((1058-1)^5\), every term of the expansion has the factor 1058 except \((-1)^5=-1\)
So \(1057^5\) is also 1 less than a multiple of 23, and adding 1 makes it divisible
(C) Unit digit of \(6^{15}-7^4\) — this matches (I) 5.
Every power of 6 ends in 6, so \(6^{15}\) ends in 6
\(7^4=2401\), which ends in 1
\(6^{15}\) is far larger than \(7^4\), so the difference ends in \(6-1=5\)
(D) Find the product of any number and the 1st whole number — this matches (III) 0. The whole numbers are 0, 1, 2, 3, …, so the first whole number is 0, and any number multiplied by 0 gives 0.