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Math Help - Find last two digit of a number

  1. #1
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    Find last two digit of a number

    Find the last two digits of  3^{83} + 21^{401}

    so:

     3^{83} \equiv M      (mod 100)
     21^{401} \equiv N      (mod 100)

    So last two digits is M+ N

    I dont really know how to continue...

    Any help would be appreciated
    Thanks in advance
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  2. #2
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    By calculating 3^{20} and 21^5 we can see that

    3^{20} \equiv 1 \mod 100 and

    21^5 \equiv 1 \mod 100.

    Therefore,

    3^{83} = 3^{80} \cdot 3^3 \equiv 3^3 \mod 100

    21^{401} = 21^{400} \cdot 21 \equiv 21 \mod 100

    27 + 21 = 48 \equiv 48 \mod 100

    Answer is 48.
    Last edited by icemanfan; February 16th 2010 at 01:28 PM. Reason: fixed mistake
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  3. #3
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    Quote Originally Posted by icemanfan View Post
    27 \cdot 21 = 567 \equiv 67 \mod 100

    Answer is 67.
    thanks

    But answer should be 27 + 21 = 48 ?

    Is there any simpler way to compute this without calculating 3^20?
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  4. #4
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    Quote Originally Posted by firebio View Post
    thanks

    But answer should be 27 + 21 = 48 ?

    Is there any simpler way to compute this without calculating 3^20?
    You are correct, I made a silly mistake at the end. If you can determine that 3^{20} \equiv 1 \mod 100 without actually calculating 3^{20}, then yes. Personally, I don't know how you would do it. In most of these types of problems, the cycle is much shorter than 20.
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    You are correct, I made a silly mistake at the end. If you can determine that 3^{20} \equiv 1 \mod 100 without actually calculating 3^{20}, then yes. Personally, I don't know how you would do it. In most of these types of problems, the cycle is much shorter than 20.
    i see.. i asked because this was a test question and calculating 3^20 seems a bit harsh.

    Thanks so much.
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