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Question on Number theory

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data

    How many four-digit numbers formed of only odd digits are divisible by five?

    2. Relevant equations

    Permutations

    3. The attempt at a solution

    Here is what I think should be done :

    Ans : 4P3 * 1
    = 24

    Is that right ?
     
  2. jcsd
  3. Jan 30, 2013 #2

    HallsofIvy

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    You don't say how you got that number so I don't see any way to comment except to say that 4P3= 4 is clearly NOT the correct answer. You don't say why you think that is true. How did you get that?

    There are a total of 4 digits in the number and 5 odd digits. How many choices are there for the first digit? The second ? The third? The fourth?
     
  4. Jan 30, 2013 #3

    tiny-tim

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    why? :confused:
     
  5. Jan 30, 2013 #4
    here is how I did it :

    The last digit is reserved for "5" since we want it to be divisible by "5"

    Then, the choices for the first digit are : 4
    2nd digit : 3
    3rd digit : 2
    Ergo, 4P3 * 1 = 4P3 = 24

    I am honestly surprised why this method is incorrect .
     
  6. Jan 30, 2013 #5

    tiny-tim

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    ah, you're assuming they all have to be different :redface:

    they don't :smile:

    (btw, i'm not familiar with this 4P3 notation, but it doesn't look right …

    24 = 4!, so where does 3 come into it? :confused:)
     
  7. Jan 30, 2013 #6
    Hmmm...I think you are right, I must not assume this.

    Honestly, I have no experience how to calculate the permutations if digits can be repeated.

    I'll try anyway :

    There are 5 choices for each of the first three digits and one choice for the last digit.
    5*5*5*1 = 125
     
  8. Jan 30, 2013 #7

    Dick

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    Right!
     
  9. Jan 30, 2013 #8

    tiny-tim

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    :biggrin: Woohoo! :biggrin:
     
  10. Jan 31, 2013 #9

    HallsofIvy

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    If the digits can be repeated, this is NOT a permutations problem. It is simply a application of the "fundamental counting principle": if A can be done in m ways and B can be done, independently, in n ways the A and B can be done in mn ways.
    There are 5 ways to choose the first digit, 5 ways to choose the second digit, 5 ways to choose the third digit, and only one way to choose the last digit which u must be 5.
    5(5)(5)(1)=

    Exactly right.
     
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