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Math Help - Lagrange's theorem

  1. #1
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    Lagrange's theorem

    Z6={1,2,3,4,5,} (modulo 6) the action * - is multiply. n=|Z|
    By Lagrange's theorem - for every number that belong to Z6 (lets say x). so x^n sepopse to be the neutral number. so x^n sepose to be 1.
    but 5^5=3125 -> 3125/6=3120 -> 3125-3120=5 -> 5^5 (modulo 6) ≠ 1
    same with the other numbers....

    so what am i doing wrong???
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  2. #2
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    Quote Originally Posted by tukilala View Post
    Z6={1,2,3,4,5,} (modulo 6) the action * - is multiply. n=|Z|
    By Lagrange's theorem - for every number that belong to Z6 (lets say x). so x^n sepopse to be the neutral number. so x^n sepose to be 1.
    but 5^5=3125 -> 3125/6=3120 -> 3125-3120=5 -> 5^5 (modulo 6) ≠ 1
    same with the other numbers....

    so what am i doing wrong???
    Z_6 = \{0,1,2,3,4,5\}

    order is 6 ... 5^6 = 15625 \equiv 1(mod \, 6)
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  3. #3
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    ok
    and what about Z6 /{0}?
    so now it's have 5 elements
    Z6/{0} = {1,2,3,4,5}
    order is 5... 5^5 (modulo 6) ≠ 1
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    \mathbb{Z}_6^\times is not a group under multiplication. There are no inverses to 2 and 3.

    \mathbb{Z}_n^\times is a group if and only if n is prime. In that case it has order p-1 and indeed you can check that a^{p-1}\equiv 1 \mod p for all a not divisible by p. This is what Fermat's "little theorem" says; it's actually a very specific case of Lagrange's theorem.
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  5. #5
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    lets say that i need to find the last 3 digit of the number: 998^1010
    i know that to find the last 3 digits, i need to use modulo 1000
    so im looking at the group Z1000 under multiplication, right?
    and now: 998^1000 = 998^(999+11) = (998^999)*(998^11)
    is it true to say that 998^999 = 1 (by Lagrange's theorem)?? if yes, so why in the group Z6/{0} = {1,2,3,4,5} under multiplication 5^5≠1 (mod 5). 1000 is not a prime number, so why in this case Lagrange's throrem works?

    any way, if its true to say that( and i dont understant yet why) so 998^11 = -2^11 = 1024*(-2) = -48 = 952 (mod 1000)
    so the last 3 numbers are:952

    so when i can use Lagrange's theorem and when not?
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by tukilala View Post
    lets say that i need to find the last 3 digit of the number: 998^1010
    i know that to find the last 3 digits, i need to use modulo 1000
    so im looking at the group Z1000 under multiplication, right?
    and now: 998^1000 = 998^(999+11) = (998^999)*(998^11)
    is it true to say that 998^999 = 1 (by Lagrange's theorem)?? if yes, so why in the group Z6/{0} = {1,2,3,4,5} under multiplication 5^5≠1 (mod 5). 1000 is not a prime number, so why in this case Lagrange's throrem works?

    any way, if its true to say that( and i dont understant yet why) so 998^11 = -2^11 = 1024*(-2) = -48 = 952 (mod 1000)
    so the last 3 numbers are:952

    so when i can use Lagrange's theorem and when not?
    Did you read my post? That is not a group.
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tukilala View Post
    998^1000 = 998^(999+11) = (998^999)*(998^11)
    is it true to say that 998^999 = 1 (by Lagrange's theorem)??
    998^{999} \equiv 312 mod 1000...as Bruno said, using multiplication modulo n, Lagrange's theorem holds if and only if n is prime. In this case we call it "Fermat's Little Theorem".

    I always double check my working in such problems with a calculator. Here the numbers are big so I used Maple, but you could try either Wolfram Alpha or the calculator function on google (although I couldn't get google to work for this example).
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  8. #8
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    Wolfam Alpha is nice..
    but when im doing a test, i dont have this web site to calculate for me those numbers...
    so how can i calculate
    998^999 (mod 1000)
    without internet
    ??
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  9. #9
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tukilala View Post
    but when im doing a test, i dont have this web site to calculate for me those numbers...
    No, but the "trick" is to get so well versed in such questions as to, eventually, not need such a program.

    Quote Originally Posted by tukilala View Post
    so how can i calculate
    998^999 (mod 1000)
    without internet
    ??
    I think the answer to this is: not easily.

    Note that 1010 is even and so 2^{1010} = (-2)^{1010}.

    Hint: prove that 2^3(2^{100}-1) \equiv 0 mod 1000 (note that 24=3*8).

    The problem with this solution is that this hint was not easy to either "see" or prove without a calculator.

    There may be a more elegant answer, but that's the best I can conjure up at the moment.
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