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The number of positive integral solutions is to be found

  1. Apr 8, 2013 #1

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    I am preparing for an entrance and I came across this sum.The equation given is xyz=3000. we need to find how many positive integral solutions are there for x,y and z.
    Please help.
     
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  3. Apr 8, 2013 #2
    Let's factor 3000 into primes.
    We are left with 23*3*53.

    Now we are left with a very simple combinatorics question. Can you figure it out? Think about what items you are selecting and what you are placing them into.
     
  4. Apr 8, 2013 #3

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    I already tried that.x can be 2^0,2^1,2^2or2^3 that is 4 ways ,2ways for 3and 4ways for 5.But what about y?
    I thought of another method ,
    using AM>GM,

    x+y+z>=43 and max can be 3002 (when one of them is 3000 nd the other two 1 each.)
    Then use multinomial theorem.Can this be done?
     
  5. Apr 9, 2013 #4
    You are vastly overthinking this.
    Consider the following problem: we have 3 red balls, 3 blue balls, and a green ball. How many different ways can we allocate them between 3 buckets?
     
  6. Apr 9, 2013 #5

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    Oh!...then the answer would be
    9P7/(3!)^2.
    thanks a lot!!!
     
  7. Apr 9, 2013 #6

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    Oh!...then the answer would be
    9P7/(3!)^2.
    thanks a lot!!!
     
  8. Apr 9, 2013 #7

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    Supposing we were to find all integral solutions???
     
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