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Proof: divisibility of 5

  1. Nov 14, 2007 #1
    I need some help proving this statement.

    Prove that a positive integer is divisible by 5 if and only if it's last digit is either
    0 or 5.

    Thanks
     
  2. jcsd
  3. Nov 14, 2007 #2
    I need some help proving this statement.

    1. The problem statement, all variables and given/known data

    Prove that a positive integer is divisible by 5 if and only if it's last digit is either
    0 or 5.

    2. Relevant equations



    3. The attempt at a solution
     
  4. Nov 14, 2007 #3

    CompuChip

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    One way is easy. Suppose n is a positive integer that ends in 0 or 5. Then we can write
    [tex]n = 10k + 5\epsilon[/tex]
    with k a positive integer and [itex]\epsilon[/itex] = 0 or 1. For example,
    1234985135 = 123498513 * 10 + 5. Then obviously [itex]n / 5 = k + \epsilon[/itex] which is a positive integer again.

    Also this approach should give you a clue for the other direction (suppose n is divisible by 5, then you can write it as 5k for some positive integer k. Now what can you say about k?)
     
  5. Nov 14, 2007 #4

    CompuChip

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    Let's continue the conversation here, so we don't have to double post as well.
     
  6. Nov 15, 2007 #5

    HallsofIvy

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    Better yet, I've merged the two threads- into this one since the problem doesn't seem to me to have a lot to do with "Computer Science and Technology"!
     
  7. Dec 17, 2007 #6
    every integer in the decimal system can be written as follows:

    [tex]z = a_0 + a_1*10 + a_2*10^2 + ... + a_n*10^n[/tex]

    as 10 is the product of 2 and 5 ==> 5 | 10

    for n > 0 all terms have 0 as the last digit

    if z is a number with last digit = 0, then [tex] a_0 = 0[/tex] ==> 5 | z

    else [tex] a_0 = 5[/tex] and also this implies 5 | z
     
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