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Math Help - What would the last 2 digits of 2008^2008 be?

  1. #1
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    Lightbulb What would the last 2 digits of 2008^2008 be?

    Hello,
    My question is as stated in the title.
    Does anyone know an easy way to solve this.
    It doesn't show the answer on a calculator and I don't want to always rely on Wolfram Alpha
    Any help would be appreciated.
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  2. #2
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    Re: What would the last 2 digits of 2008^2008 be?

    Quote Originally Posted by Atrey_G View Post
    Hello,
    My question is as stated in the title.
    Does anyone know an easy way to solve this.
    It doesn't show the answer on a calculator and I don't want to always rely on Wolfram Alpha
    Any help would be appreciated.
    Examine only the last digit:

    (2008)^1 = xxxx...xxxx8
    (2008)^2 = xxxx...xxxx4
    (2008)^3 = xxxx...xxxx2
    (2008)^4 = xxxx...xxxx6
    ----------------------------
    (2008)^5 = xxxx...xxxx8
    ...

    Do you see the pattern?
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  3. #3
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    Re: What would the last 2 digits of 2008^2008 be?

    we want to find 20082008 (mod 100)

    note that this will be the same as 82008 (mod 100).

    (since 2008 = 8 (mod 100)).

    now 821 = (83)7 = 127 = (122)3(12) (mod 100)

    = (443)(12) = (442)(44)(12) = (36)(44)(12) = (84)(12) = 8 (mod 100).

    since 2008 = (21)(95) + 13:

    82008 = (821)95(813) = (895)(813) (mod 100)

    and 95 = (21)(4) + 11:

    (895)(813) = (821)4(811)(813) (mod 100)

    = 84+11+13 = 828 = (821)(87) = (8)(87) (mod 100)

    = 88 = (83)2(82) = (12)2(64) = (44)(64) = 16 (mod 100).

    so the last two digits are 16.
    Thanks from Goku and Atrey_G
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  4. #4
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    Re: What would the last 2 digits of 2008^2008 be?

    Just in case (like me) you are not comfortable with all the modular arithmetic,
     2008^{2008}
     = (2000+8)^{2008}
     = 2000C0 + 2000^{2008} + 2008C1 (2000^{2007})(8^1)+ ..........+2008C2008 (8^{2008})
    Now except the last term, the rest of the terms are going to have atleast 3 Zeroes in the end. So the last two digit of the expression will be decided by the last expression, namely  8^{2008} alone.
     8^{2008} = 2^{6024}
    Now we just have to see the pattern in powers of 2. So it comes out as 2,4,8,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44 ,88,76,52,04,08,16 ..
    What we see is the the last digit is 6 for all multiples of 4 and that the last 2 digit follows a pattern namely - 16,56,96,36,76,16 ... (for powers in multiple of 4) , so for  2^4, 2^{24} etc, last digit is 16. Now we have  2^{6024} and  6024=24*251
    Hence the last two digits are 16.
    Thanks from Atrey_G
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