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 \(61\times62\times63\times64\times\ldots\times69\) 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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
The unit’s digit of a product or a sum depends only on the unit’s digits of the numbers. The unit’s digits of the powers of 7 repeat in a cycle of four: 7, 9, 3, 1. Any power of a number ending in 6 ends in 6.
The two powers used in (A) and (D):
\(153=4\times38+1\), so \((257)^{153}\) ends like \(7^1\), in 7
\((346)^{72}\) ends in 6
(A) Unit’s digit of \((257)^{153}\times(346)^{72}\) — this matches (IV) 2. \(7\times6=42\), which ends in 2.
(B) Unit’s place of the product \(61\times62\times63\times\ldots\times69\) — this matches (III) 0. The product contains 65 and an even number such as 62, and 5 times an even number ends in 0.
(C) 121012 divided by 12, the remainder — this matches (II) 4. \(12\times10084=121008\), and \(121012-121008=4\).
(D) Unit’s digit of \((257)^{153}+(346)^{72}\) — this matches (I) 3. \(7+6=13\), which ends in 3.