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Mod question I thought I got but I don't.

  1. Feb 24, 2012 #1
    For n in N (Natural Numbers), use modular arithemetic to establish each of the following divisibility statements:

    a) 7 | 52n + 3 ( 25n-2)

    Attempt: Well I've set up a mod 7 table and was figuring out the values from n = 0,....6 for the two individual expressions, then I was going to combine them. But I of course hit a rough patch when I got to 25n - 2. Since I'm in the Natural numbers, fractions do not exist, so how do I treat 25n - 2 when n = 0 ? (Note: This is what I attempted prior to getting help originally, then I was told that 0 would not be included in the Natural numbers for this question.)......but I was working through this question but it occurred to me that I'm trying to show that 7 | 52n + 3 ( 25n-2), so wouldn't I have to show that:

    52n + 3 ( 25n-2) "congruent" 0 (mod 7)?.......but how can I show this if 0 is not suppose to be included in my work?

    A follow up to that question is that the question asked me to establish that the above expression is true. Is the only way I can establish its truth is by using a mod table or is there another method? and does this apply to most questions of this nature?
     
  2. jcsd
  3. Feb 24, 2012 #2

    SammyS

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    You need to think more specifically about what can be 0 and what can't be 0 in this problem.

    The problem states explicitly "For n in N (Natural Numbers)". This has nothing to do with having a 0 in
    52n + 3 ( 25n-2) ≣ 0 (mod 7) .​
    Having a 0 is OK in this expression. It's the quantity, n, which is to come from the natural numbers !

    It appears that this statement can be solved inductively.
     
  4. Feb 24, 2012 #3

    Deveno

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    n does not have to be in the set {0,1,2,3,4,5,6}.

    n can be any natural number ≥ 1 (the original statement clearly does not hold for n = 0).

    questions of this nature do not "have" to be proved via modular arithmetic, induction, or other forms of logical proof are just as valid, but in this particular case, modular arithmetic makes the proof EASY (we don't even have to calculate any "test values").
     
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