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Playing around with the modulo operation

  1. Jun 18, 2010 #1
    Hey guys,
    I am by no means a mathematician, but I do have a decent eye for patterns, and I found a pretty cool one today. I was hoping one of you guys could tell me more about it.

    As a general rule, I've found that if (a*b) mod c = 1, then the sequence a^n mod c is the reverse of the sequence b^n mod c.

    For example, (7*8) mod 11 = 1
    and 7^n mod 11= 7,5,2,3,10,4,6,9,8...
    while 8^n mod 11= 8,9,6,4,10,3,2,5,7...

    As another example, (56*24) mod 17 =1
    and 56^n mod 17 = 5,8,6,13,14,2,10,16,12,9,11...
    while 24^n mod 17= 11,9,12,16,10,2,14,13,6,8,5...

    What do you all think about this? I'm willing to bet that all I've done is show a simple concept in a convoluted way, but I'm to fried to think critically about this any more.
  2. jcsd
  3. Jun 18, 2010 #2


    Staff: Mentor

    This is very interesting. For your first example, it is significant that 7 and 11 are relatively prime (no factors in common other than 1), and the same is true for 8 and 11.
  4. Jun 19, 2010 #3


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    Hi Archosaur. Here's a partial proof that's very straight forward. (BTW, to save typing please take all equalities as " = modulo c" and all variables as positive integer in what follows).

    1. Assume that we have some power of "a" that is equal to some power of "b", that is [itex] a^p = b^q = x [/itex] for some p,q. (I haven't shown under what conditions this is guaranteed to occur, so this is definitely only a partial proof).

    2. Since [itex]ab = 1[/itex] then [itex]abx = x[/itex],

    so [itex] ab b^p = a^p[/itex] and hence [itex]b^{p+1} = a^{q-1}[/itex], which proves the reversed nature of the two series.

    Perhaps someone else will be able to provide a more complete proof.
  5. Jun 23, 2010 #4
    Wow, thank you very much, uart! That was just the kick-start I needed to think about this more.
  6. Jun 25, 2010 #5


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    All modulo c: If ab = 1, then a= b^-1. So a^n=b^-n. The periodicity of the sequences assures that they are opposite. a^n = b^-n is going in the "opposite direction" to b^n.
  7. Jun 25, 2010 #6
    Awesome, yes, this is what I ended up with. Thanks for your help, everyone!
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