Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Remainder when \(17^{35234}\) is divided by 8 | (I) | 0 |
| (B) | Remainder when 4444 is divided by 9 | (II) | 1 |
| (C) | Unit's digit of \((34)^{15}+(34)^{16}\) | (III) | 2 |
| (D) | Unit digit of \(7^4-9^3\) | (IV) | 7 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Remainders and units digits are found without working out the full numbers: only the remainder of the base, or its last digit, decides the result.
(A) Remainder when \(17^{35234}\) is divided by 8 — this matches (II) 1. \(17=2\times8+1\) leaves remainder 1, so every power of 17 leaves \(1^{35234}=1\).
(B) Remainder when 4444 is divided by 9 — this matches (IV) 7. A number leaves the same remainder on division by 9 as the sum of its digits: \(4+4+4+4=16\), and \(16=9+7\).
(C) Unit’s digit of \((34)^{15}+(34)^{16}\) — this matches (I) 0. Powers of a number ending in 4 end in 4 for an odd power and 6 for an even power, so the units digits are 4 and 6, and \(4+6=10\) ends in 0.
(D) Unit digit of \(7^4-9^3\) — this matches (III) 2. \(7^4=2401\) and \(9^3=729\), so \(2401-729=1672\), which ends in 2.