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Math Help - Proof

  1. #1
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    Proof

    I can't find the topic I made.. ugh.

    Any way I'm trying to prove 5^{2008} always ends in the digits 25.

    I almost had it.. something like:

    5(2x+5) = 10x + 25 . . . this doesn't quite work. I have to tweek that somehow to show it.
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  2. #2
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    Anyone? I know the proof is similar to what I have. I just can't tweek it to get integers x = 1, 2, ... to satisfy that and include all the powers of 5 to show that you get something (something like 1000 + 100 + 25) and hence will always have 00 + 25 which completes the proof.
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  3. #3
    Moo
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    Hello,

    You can prove that 5^x always ends by 25 for x \ge 2

    Do it by induction if you want to
    ------------------------------------------------
    It's ok for x=2

    Assume 5^x ends up by 25, that is to say 5^x=100k+25, where k is a (positive) integer.

    5^{x+1}=5 \cdot 5^x=5 \cdot (100k+25)=100 \cdot (5k)+5 \cdot (20+5)=100 \cdot (5k)+100+25

    5^{x+1}=100 \cdot (5k+1)+25, which obviously ends up by 25
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