Question:
What is the remainder when \(4^{96}\) is divided by 6?
What is the remainder when \(4^{96}\) is divided by 6?
Show Hint
Look at the remainders of 4, 4\(^2\), 4\(^3\), ... when divided by 6 and see whether a pattern repeats.
Updated On: Nov 6, 2026
- 0
- 4
- 3
- 2
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The Correct Option is B
Solution and Explanation
Step 1: Remainders of small powers:
4 divided by 6 leaves remainder 4. \(4^2 = 16\) leaves remainder 4. \(4^3 = 64\) leaves remainder 4 (since 64 = 6 x 10 + 4).
Step 2: Show it never changes:
If \(4^k\) leaves remainder 4, then \(4^{k+1}\) leaves the remainder of \(4\times 4 = 16\), which is 4. So every power \(4^k\) (k at least 1) leaves remainder 4.
Step 3: Conclude:
Therefore \(4^{96}\) leaves remainder 4 when divided by 6. This is option (2).
Final Answer:
The correct option is (2): 4
4 divided by 6 leaves remainder 4. \(4^2 = 16\) leaves remainder 4. \(4^3 = 64\) leaves remainder 4 (since 64 = 6 x 10 + 4).
Step 2: Show it never changes:
If \(4^k\) leaves remainder 4, then \(4^{k+1}\) leaves the remainder of \(4\times 4 = 16\), which is 4. So every power \(4^k\) (k at least 1) leaves remainder 4.
Step 3: Conclude:
Therefore \(4^{96}\) leaves remainder 4 when divided by 6. This is option (2).
Final Answer:
The correct option is (2): 4
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