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Simple math question.

  1. Mar 9, 2009 #1
    The sum of four numbers equals the product of four numbers equals 711.

    What are the four numbers?

    I see 2 equations with 4 unknowns, and no way to do this but brute force it; is there a special circumstance here?
     
  2. jcsd
  3. Mar 9, 2009 #2
    The sum and product of the same 4 numbers are equal to 711? Is that what you mean?

    The prime decomposition of 711 is 3^2*79. There aren't too many ways to get 4 numbers; you either have 1, 1, 9, 79, or 1, 1, 3, 237 or 1, 3 , 3, 79. None of those sum up to 711. Can you restate the problem more clearly?
     
  4. Mar 9, 2009 #3
    Admittedly, he did say "numbers" and not "integers", so the prime factorization of 711 isn't necessarily relevant, although I agree it's certainly implied. Hmm...

    DaveE
     
  5. Mar 9, 2009 #4
    Hm...Just thinking out loud here. This approach requires making certain assumptions that could prove to be wrong. Please forgive my formatting errors.

    711 = a+b+c+d
    711 = abcd

    The first two assumptions are that a = 711, so:

    0 = b+c+d
    1 = bcd

    The next one is that bc = 1, which means that

    d = -1

    and

    -1 = bc

    and that

    -1/b = c

    which means that

    711 = 711-1+(-1/b)+b
    0 = -1-(1/b)+b
    0 = b^2-b-1
    b = [1+-(1-4*-1*1)^1/2]/2
    b = [1+-(5)^1/2]/2
    b = 1.618 or -.618

    Therefore, since -1/b = c

    When b = 1.618, c = -.618 and when b = -.618 c = 1.618
    And that since addition and multiplication are communicative, one of these solution is redundant.

    So, in conclusion (finally :smile:)

    Since -1 = bc

    711 ?= 711*-1*-1
    711 = 711

    and

    711 ?= 711-1-.618+1.618
    711= 711

    a = 711
    b = -.618
    c = 1.618
    d = -1

    Q.E.D.
     
    Last edited: Mar 9, 2009
  6. Mar 10, 2009 #5
    As the op said, these are two equations in 4 unknowns. There are infinitely many solutions. F(x) has added 3 more equations which seems extravagant under the circumstances.
     
  7. Mar 10, 2009 #6
    Haha. Nevertheless, those three extravagant equations allowed me to find one possible solution.

    Oh, and by the way, prove that there are a infinite number of solutions.
     
    Last edited: Mar 10, 2009
  8. Mar 10, 2009 #7
    That's pretty easy, no?

    w+x+y+z = 711
    wxyz = 711
    w = 711/xyz

    Assume z=1:

    w=711/xy
    711/xy+x+y+1 = 711
    711 + x^2y + xy^2 = 710xy
    xy^2+(x^2-710x)y+711 = 0

    You can use the quadratic formula (though it's messy) to solve for y, with:

    A = x
    B = x^2-710x
    C = 711

    So as long as 2A =/= 0, and B^2 > 4AC, you're good. Specifically:

    x =/=0
    x^4+2x^2-1420x+504100x^2 > 2840x
    x^3+2x+504100x > 4260

    So, without bothering to solve it, it's obvious that for x>1, the quadratic will be solvable, because it'll satisfy the above requirements. And obviously there are an infinite number of values for x>1.

    DaveE
     
  9. Mar 10, 2009 #8
    My bad. f(x) only added two equations, the expected number. As not both of the original equations are linear it is not enough to merely count equations and unknowns as I did. However, davee123 has saved the day for me on this point.
     
  10. Mar 11, 2009 #9
    Haha fine. I'm too tired to look through your post right now, so I'll just agree for the moment.

    Just as a side note, my math teacher has a problem that is exactly similar to this one that that he calls the "7-11 Problem" (as in the gas station). When asked, he said that there was only 1 possible solution to this, and I am tempted to agree because the problem combines addition and multiplication. But, he is not infallible.
     
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