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Math Help - Order

  1. #1
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    Order

    Can someone show why the order of 10 is 11 while considering the group: Integers mod 55

    thanks
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  2. #2
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    Re: Order

    presumably you mean the integers mod 55 under addition.

    in this group, a "power" of 10, 10^k, is 10 added k times:

    10^2 = 10 + 10
    10^3 = 10 + 10 + 10
    (etc.)

    so we will write 10^k as k*10.

    the most direct route is to show that k = 11, is the smallest positive k that will work.

    1*10 = 10
    2*10 = 20
    3*10 = 30
    4*10 = 40
    5*10 = 50
    6*10 = 5 (remember, we are working mod 55)
    7*10 = 15
    8*10 = 25
    9*10 = 35
    10*10 = 45
    11*10 = 0 (110 = 0 mod 55, since 110 = (55)(2) + 0).
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  3. #3
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    Re: Order

    Thank! One more thing. What would the order of 0 and 1 and 5 be while looking at the integers mod 6?
    Order(2) = 3
    Order(3) = 2
    Order(4) = 3
    But I'm not sure about 0,1 and 5
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  4. #4
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    Re: Order

    why don't you try and calculate?

    for 0, what is:

    0
    0+0
    0+0+0
    0+0+0+0
    0+0+0+0+0
    0+0+0+0+0+0

    for 1, what is:
    1
    1+1
    1+1+1
    1+1+1+1
    1+1+1+1+1
    1+1+1+1+1+1

    for 5, what is
    5
    5+5
    5+5+5
    5+5+5+5
    5+5+5+5+5
    5+5+5+5+5+5?
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  5. #5
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    Re: Order

    So,
    Order(1) = 6
    Order(5) = 6
    Order(0) = 0?
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  6. #6
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    Re: Order

    very close....the order of 0 is 1 (the order is always a positive number). you got the other two right.
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