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Math Help - remainder of sum, divided by 4

  1. #1
    Member cassiopeia1289's Avatar
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    remainder of sum, divided by 4

    Ok, so I did the problem and got a solution that another classmate didn't get. Now its got me paranoid, I don't know whose right.

    "What is the remainder when the following sum is divided by 4?": 1^5 + 2^5 + 3^5 + ... + 99^5 + 100^5

    So I'll briefly explain - I did:
    so that is \equiv 25(1^5 + 2^5 + 3^5 + 0^5) (mod 4)
    worked it all out so that it is now congruent to 1(1 + 0 + 0) (mod 4)
    Thus, the remainder is 1, yes??

    My classmate go that there was no remainder. Did I do something wrong in calculations??
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  2. #2
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    Quote Originally Posted by cassiopeia1289 View Post
    Ok, so I did the problem and got a solution that another classmate didn't get. Now its got me paranoid, I don't know whose right.

    "What is the remainder when the following sum is divided by 4?": 1^5 + 2^5 + 3^5 + ... + 99^5 + 100^5

    So I'll briefly explain - I did:
    so that is \equiv 25(1^5 + 2^5 + 3^5 + 0^5) (mod 4)
    worked it all out so that it is now congruent to 1(1 + 0 + 0) (mod 4)
    Thus, the remainder is 1, yes??

    My classmate go that there was no remainder. Did I do something wrong in calculations??
    Your idea is very good, but be careful, 3^5\equiv(-1)^5\equiv -1\equiv 3\,({\rm mod}\,4) hence 1^5+2^5+3^5+4^5\equiv 1+0+3+0\equiv 0\,({\rm mod}\,4), and your classmate is right.

    By the way, 3^5\equiv 0\,({\rm mod}\, 4) would mean that 4 divides 3^5. Of course, this can't be.
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  3. #3
    Member cassiopeia1289's Avatar
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    oh ok - thanks! I knew it was something stupid I did!
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