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Quick questions about modular arithmetic

  1. May 24, 2005 #1
    [tex]99999^{99} + 1[/tex]
    As [tex]99999 \equiv[/tex]24[tex](mod \25)[/tex]
    Can I say then:
    [tex]99999^{99} + 1 \equiv[/tex]24[tex]^{99} + 1(mod \25)[/tex], or is it
    [tex]99999^{99} + 1 \equiv[/tex]24[tex]^{99}(mod \25) + 1[/tex],
    or are these two the same things?
  2. jcsd
  3. May 24, 2005 #2


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    This one is standard. People would probably understand the second, but the usual format is to write the mod at the end.
  4. May 24, 2005 #3
    oh ok, i was just wondering if the two were the same. I was uncertain if by moving the +1 before the mod5 it would change the meaning. Awesome, well then that makes this question a lot easier now. Thanks :D
  5. May 25, 2005 #4
    Another quick question.

    Are the following two congruencies the same?
    1. [tex]24^{99} + 1 \equiv 0 \mod \25[/tex]
    2. [tex]24^{99} + 1 \mod \25 \equiv 0 \mod \25[/tex]

    I am trying to show that a number is divisible by 25, and I found that 25 can be written in terms expressed in equation #1. And I found that number that I am trying to divide by 25 can be expressed in terms expressed in equation #2.

    I thought I finished the proof but now that I am looking at it, I am unsure about this one thing.
  6. May 25, 2005 #5


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    When you write 25 don't write it \25 or the 5 just shows up, write it normally, e.g:

    [tex]24^{99} + 1 \equiv 0 \mod 25[/tex]


    [tex]24^{99} + 1 \mod 25 \equiv 0 \mod 25[/tex]

    There two statements are the same, I think you miss the point though. Something like:

    [tex]24^{99} + 1 \mod 37 \equiv 0 \mod 25[/tex]

    Woule make not really make that much sense, so amoung other reasons there is no reasons to write the mod twice.
  7. May 25, 2005 #6
    Okay. Because I was trying to show that equation #2 and #1 are equivalent.

    So they are right? LOL sorry, I'm just very uncertain about myself.
  8. May 25, 2005 #7
    Well, people tend to view congruence defined as a ternary relation:

    [tex]x \equiv y \mod n \overset{def}{\Longleftrightarrow} n|x-y.[/tex]

    Sometimes one omits the modulo part, but it is still understood that we're dealing with modulo arithmetic by using the equivalence sign [tex]\equiv[/tex], instead of an equality sign. Another way of stating a congruence [tex]x \equiv y \mod n[/tex] is by saying that [tex]x[/tex] and [tex]y[/tex] belong to the same residue class (look this up on mathworld.wolfram.com). That is

    [tex]x \equiv y \mod n \Leftrightarrow [x]_n = [y]_n[/tex]

    So, your equation 2. states (with the missing 2 from 25) that

    [tex][[25^{99}+1]_{25}]_{25} = [0]_{25}[/tex]
  9. May 26, 2005 #8
    Remember Matt Grime's post in your other thread. You are trying to show that
    24^99 +1 = 0 mod 25.
    but 24^99 +1 = (-1)^25 +1 = -1+1 = 0 mod 25, since 24=-1 mod 25 and since -1 raised to an odd power is -1. Q.E.D.
  10. May 26, 2005 #9
    Yea, thats exactly how I did it Ramsey :D Sorry, I should of concluded this thread by mentioning that. Thanks for another helpful response though. I appreciate it.
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