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Math Help - Product of factors

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    Product of factors

    In how many ways can 2310 be expressed as a product of 3 factors ? ( This will not involve Combinatorics analysis !)
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    Re: Product of factors

    Hello, SheekhKebab!

    In how many ways can 2310 be expressed as a product of 3 factors?
    (This will not involve Combinatorics analysis!) . It won't?

    I suppose a brute-force LIST is acceptable . . .

    . . \begin{array}{c}2\cdot3\cdot385 \\ 2\cdot5\cdot231 \\ 2\cdot7\cdot165 \\ 2\cdot11\cdot105 \\ 3\cdot5\cdot154 \\ 3\cdot7\cdot110 \\ 3\cdot11\cdot70 \\ 5\cdot7\cdot66 \\ 5\cdot11\cdot42 \\ 7\cdot11\cdot30 \end{array}

    Answer: 10 ways.
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    Re: Product of factors

    Quote Originally Posted by Soroban View Post

    [size=5]
    I suppose a brute-force LIST is acceptable . . .
    Hi Sir,

    long time, no see ! There must be some algebraic method to solve this without Combinatorics, which is also simple and easy. And by the way the correct answer is 41.
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    Re: Product of factors

    Quote Originally Posted by SheekhKebab View Post
    In how many ways can 2310 be expressed as a product of 3 factors ? ( This will not involve Combinatorics analysis !)
    Restating the problem

    Find all {x,y,z} such that x,y,z integers, xyz = 2310, and 1\leq  x \leq  y \leq  z

    The number of solutions is 41
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    Re: Product of factors

    Quote Originally Posted by SheekhKebab View Post
    Hi Sir,

    long time, no see ! There must be some algebraic method to solve this without Combinatorics, which is also simple and easy. And by the way the correct answer is 41.
    It looks comprehensive enough to me. Are you sure that you are't counting solutions like 11 x 5 x 42? If so then you do need the combinitorics after all.

    -Dan
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    Re: Product of factors

    You've forgotten one of factors may be 1
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    Re: Product of factors

    Quote Originally Posted by Cartesius24 View Post
    You've forgotten one of factors may be 1
    also 11,14,15 is not in the list
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    Re: Product of factors

    Quote Originally Posted by Idea View Post
    Restating the problemFind all {x,y,z} such that x,y,z integers, xyz = 2310, and 1\leq  x \leq  y \leq  zThe number of solutions is 41
    Hi Idea,

    Please explain your method. How did you get to 41 ?
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    Re: Product of factors

    Quote Originally Posted by SheekhKebab View Post
    Hi Idea,

    Please explain your method. How did you get to 41 ?
    same as Soroban brute force list except I included 1 as a possible factor, and the first two factors don't have to be prime numbers

    such as 6 11 35 and 7 10 33 are included in the list

    There are 16 triplets of the form {1,y,z} since there are 32 divisors of 2310

    I don't know how but I sense that there is a way to count the remaining 25. Is that what you are after?
    Last edited by Idea; August 14th 2014 at 03:04 AM.
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    Re: Product of factors

    Quote Originally Posted by Idea View Post
    I don't know how but I sense that there is a way to count the remaining 25. Is that what you are after?
    Yep! I think the method without combinatorics will be similar to the method we use to express a number as a product of 2 factors. Also, I would like to know what will be the method under combinatorics for this question.
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