Match the List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Number | Unit digit of the number | ||
| (A) | \(4^{425}\) | (I) | 6 |
| (B) | \(4^{1024}\) | (II) | 4 |
| (C) | \(133^{63}\times4^{49}\) | (III) | 3 |
| (D) | \(3^{81}\) | (IV) | 8 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-II, B-I, C-IV, D-III
The unit digit of a power depends only on the unit digit of the base, and it repeats in a cycle: powers of 4 end in 4, 6, 4, 6, ... (4 for an odd power, 6 for an even power); powers of 3 end in 3, 9, 7, 1 and then repeat, so the remainder of the power on division by 4 decides the digit.
(A) \(4^{425}\) — this matches (II) 4. The power 425 is odd.
(B) \(4^{1024}\) — this matches (I) 6. The power 1024 is even.
(C) \(133^{63}\times4^{49}\) — this matches (IV) 8.
\(133^{63}\) ends like \(3^{63}\); \(63=4\times15+3\), so it takes the 3rd digit of the cycle, 7
\(4^{49}\) has an odd power, so it ends in 4
\(7\times4=28\), so the product ends in 8
(D) \(3^{81}\) — this matches (III) 3. \(81=4\times20+1\), so it takes the 1st digit of the cycle, 3.