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  1. Nov 12, 2003 #1
    4 items add to $14.26 and multiply to $63.42. Have fun

    I need this solved and have finally been stumped, how is this equation solved?

    post here or email me please frank_medewar@hotmail.com

    thank you all
  2. jcsd
  3. Nov 12, 2003 #2
    please I urgently need to know how I can solve this...

    or at least a resource that can help me out

    thakn you all kindly
  4. Nov 12, 2003 #3

    Tom Mattson

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    Homework problems go in the Homework Forums.


    Chill out. This is not an instant Homework Help service. That said, you have to try the problem and show some work before we offer any assistance. Please see this thread for guidelines.

    So, show us what you have and we will show you where you are going wrong.
  5. Nov 12, 2003 #4
    thanks TOM!! much help, im not in school however

    I just love math..mainly TRIG and graphs n such--i saw the problem ,and I cannot figure it out forteh life of m, i dont even know where to start.

    but would love to know how to solve this type of equation....
  6. Nov 12, 2003 #5
    I think that this can be solved by using linear equations but it is really complicated.
  7. Nov 12, 2003 #6

    is this a joke?
  8. Nov 12, 2003 #7


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    What have you tried? you surely tried a couple things before lookig for internet help.

    Have you tried setting up a couple equations? where do things get hard?
  9. Nov 13, 2003 #8
    It seems it is too complex for the MS Aceess solver tool.
  10. Nov 13, 2003 #9
    4 items add to $14.26 and multiply to $63.42. Have fun

    Something wrong here.
    If every item is a price in $, their product will be in $^4.

    So the items better be pure numbers, positive numbers.

    Trial and error! what else?
    3,7, 101,2 are prime factors. Multiplying each of these by an integral power of 10 will yield an infinite set of factors.
    Keep trying! Best of luck!
  11. Nov 13, 2003 #10
    Restate the problem as the sum of four numbers {A,B,C,D} is 1426 and the product of four numbers {A,B,C,D} is 6342 where {A,B,C,D} are integers. Working with integers is easier. Next develope a few conclusions: 1<= {A,B,C,D} <= 1423 because if A=1, B=1, c=1 then D=1423 (this doesn't work for the product portion but it establishes boudries).

    Once you have some boundries, decompose 6342 into a product of primes {2*3*7*151}. From there you can find all factors of the product and trial and error the answer.

    A system of lineral equations would not be easy because you only have two equations and four variables (unless you're clever enough to think of two additional equations)...
  12. Nov 13, 2003 #11


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    This is not really a math problem. Linear Algebra will not get you an anwer. YOu have 4 unknowns and 2 equations, to solve this using tools like Excel solver you will need 2 more constraints.

    This shortage of infromation makes it a logic puzzle to be worked out by trial and error.

    As you say, have fun.
  13. Nov 13, 2003 #12


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    Why is this "not really a math problem"?

    It doesn't have a unique correct answer but many math problems don't.

    There are 4 unknowns with only 2 equations but, since the problem was stated in terms of money, if we convert to cents, as Faust9 did, so that a+ b+ c+ d= 1426 and abcd= 6341, we can assert that the unknowns have to be positive integers so this is a (non-linear) Diophantine problem.

    I don't claim it's easy but it doable (and certainly is a mathematics problem).
  14. Nov 14, 2003 #13
    If we do this in cents, then
    a+b+c+d = 1426, and
    abcd=1004*63.42= 6342000000.
    Now, 6342000000 = 27*3*56*7*151.

    You got to group these prime factors in 4 products so that their sum is 1426. It's a trial & error problem.
  15. Nov 14, 2003 #14
    2*2*2*2*5*5 = 400
    2*2*2*5*5 = 200
    5*5*7 = 175
    151*3 = 453
    SUM 1228

    That's too small. So, in the next try, we make the highest price bigger, to get a bigger result. And so on...
  16. Nov 14, 2003 #15
    Edit: _ means blank
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