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Math Help - Natural logs????

  1. #1
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    Natural logs????

    How would you find the remainder when 27 to the power of 1001 is divided by 13? How would you find the remainder when 38 to the power of 101 is divided by 13? Finally, How would you show that 70 times 27 to the power of 1001 plus 31 times 38 to the power of 101 is divisible by 13? Many thanks !!!!!
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  2. #2
    Super Member Matt Westwood's Avatar
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    This is obviously an elementary number theory question.

    Remember that a \equiv b \iff a^n \equiv  b^n.

    27 \mod 13 \equiv  1 \mod 13 so \frac {27^{1001}}{13} = \frac {1^{1001}} {13} and finding the answer to that shouldn't be too taxing.

    38 is slightly more difficult as 38 \mod 13 \equiv -1 \mod 13 but again it's not a brain-buster.
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  3. #3
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    ?

    Thanks Matt, but I'm 13. Can anyone plz make it simpler?

    Thanks
    Last edited by Stella; September 7th 2008 at 01:18 AM.
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  4. #4
    Super Member Matt Westwood's Avatar
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    Sorry, your age wasn't obvious from the level of mathematics you're posting in.

    The question seems to be concerned with modulo arithmetic. So I assumed you've been taught about that.

    Divide 27 by 13, you get remainder 1.

    If you square 27 and divide that by 13, you get remainder 1 squared.

    It just works like that. I'd demonstrate a proof but you're 13.

    Anyway, raise 27 to *any* number (call it n) and when you divide that by 13, you get 1 raised to the power of n.

    I won't tell you how to raise 1 to the power of n, you ought to understand about powers if you're doing this question.

    So, raise 27 to the 1001'th power, divide that by 13 and you get ...?

    Similar with 38. Divide that by 13 and you get a remainder 12, or, pushing the envelope a bit, a "remainder" of -1. (You're allowed to do that with modulo arithmetic to an extent.)

    So the same rule applies, 38 to the power of 1001, divide that by 13 and you get a remainder which is (-1) raised to the power of 1001.

    Again, I'll leave that for you to do.
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  5. #5
    Moo
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    Hello,

    I'll try to explain why !

    27 has a remainder 1 in the division by 13.
    Therefore, there exists an integer k such that 27=13k+1 (in fact, k=2, but we don't care).

    If it is multiplied by 27 again, then we would have : 27=(13k+1)*(13k+1)=13˛*k˛+2*13k+1
    We can factor 13 :
    (13k+1)(13k+1)=13(13k˛+2k) + 1
    So the remainder of 27*27 in the division by 13 is again 1 !

    More generally, if you multiply two numbers, m and n, whose remainder in the division by 13 is 1, you have :
    m=13m'+1
    n=13n'+1
    where m' and n' are integers.

    m*n=(13m'+1)*(13n'+1)=13˛*(m'n')+13m'+13n'+1
    Factor by 13 :
    m*n=13*(13m'n'+m'+n') + 1

    What is the remainder of m*n in the division by 13 ?


    It's the same reasoning for here. Each time you increase the power of 27, you multiply the previous result by 27.
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