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Bunuel
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What is the remainder when \(2^{2004}\) is divided by \(7\)?

\(2^{2004}= 2^{3*668}= 8^{668}= (7+1)^{668}\)

--> the remainder will be \(1\).

The answer is E.
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Solution



To find
We need to determine
    • The remainder, when \(2^{2004}\) is divided by 7

Approach and Working out
We know, \(2^3\) divided by 7 gives a remainder 1.
    • Hence, we can also say \(2^{3k}\) divided by 7 also gives a remainder 1.

Now, 2004 = 3 x k
Hence, Rem[\(2^{2004}/7\)] = Rem[\(2^{3k}/7\)] = 1

Thus, option E is the correct answer.

Correct Answer: Option E
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I did it in this way, correct me if i am wrong pls: 2^(4*501)/7, 2^4 gives remainder as 6 which -1 when divided by 7 and 2^501 gives remainder as 1 so i chose 1
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