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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Questions on divisibility turn on the remainder: to make a number divisible, subtract its remainder or add what is missing to reach the divisor. The units digit of a product depends only on the units digits of its factors.
(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 leaves remainder 2
\(2025-2=2023\) is divisible by 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
An odd power of a number that is 1 less than a multiple of 23 is again 1 less than a multiple of 23, so \(1057^5\) leaves remainder 22
Adding 1 makes it divisible by 23
(C) Unit digit of \(6^{15}.7^4\) — this matches (I) 5, by elimination; the printed entry does not actually match.
Every power of 6 ends in 6, so \(6^{15}\) ends in 6
\(7^4=2401\) ends in 1
So the product ends in \(6\times1=6\), not 5
(D) The product of any number and the 1st whole number is: — 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.
Note: It seems that there is a mistake in the original NTA paper: no entry of List-II fits (C). Option (2) is the only option that matches (A), (B) and (D) correctly, so (C) is left with (I). The maths gives 6, the key gives 5, and on exam day the key collects the marks.