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Math Help - Largest Product of Integers

  1. #1
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    Largest Product of Integers

    Hello, here is a problem that I came across a while ago, I have what I think is the solution, but perhaps there is an easier way to solve it, or perhaps a more precise solution.

    The problem is:

    What is the largest product of integers whos sum is 2008.

    Let me know your thoughts
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  2. #2
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    Quote Originally Posted by sean92t0 View Post
    Hello, here is a problem that I came across a while ago, I have what I think is the solution, but perhaps there is an easier way to solve it, or perhaps a more precise solution.

    The problem is:

    What is the largest product of integers whos sum is 2008.

    Let me know your thoughts
    I have approximated a lot. If it were product of reals, I believe my solution would have been fully justifiable. Let me know if my answer matches yours.

    I will assume positive integers because or else we can maximize the product without bounds.

    Lets say that there are y integers whose sum is 2008. Denote them by x_1,x_2,\cdots,x_y.

    So x_1+x_2+\cdots+x_y = 2008

    We want to maximize the product P = x_1 x_2 \cdots x_y

    Since by assumption the numbers are positive, we can apply AM-GM inequality.

    \frac{\sum_1^y x_i}{y} \geq \sqrt[y]{P} \Rightarrow P \leq \left(\frac{2008}{y}\right)^y

    Thus the maximum value of the product is \left(\frac{2008}{y}\right)^y and is purely determined by y.

    So we want to maximize f(y) = \left(\frac{2008}{y}\right)^y

    A little bit of calculus establishes that f(y) approaches maxima at y = \frac{2008}{e}

    Thus the discrete maximum is attained at y = \lfloor \frac{2008}{e} \rfloor + 1 = 739(Actually I have done trial error to choose between 738 and 739)

    In AM-GM inequality, equality holds when all the numbers are equal.

    Thus 739 x_i \approx 2008 \Rightarrow x_i =  \lfloor\frac{2008}{739}\rfloor = 3

    Thus I believe that 3^{669} is the largest product.
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  3. #3
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    this is very similiar to what I did, however I am not familiar with the inequality you used so I did that part slightly differently. You have 3^669, my problem with that is that the sum of those integers only adds up to 2007.
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    Quote Originally Posted by sean92t0 View Post
    this is very similiar to what I did, however I am not familiar with the inequality you used so I did that part slightly differently. You have 3^669, my problem with that is that the sum of those integers only adds up to 2007.
    No i have added 669 3s and one 1.
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    you're right, i read it wrong.
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    Quote Originally Posted by sean92t0 View Post
    this is very similiar to what I did, however I am not familiar with the inequality you used so I did that part slightly differently.
    The inequality is called AM-GM and you can search it on the web. Its pretty popular...

    Quote Originally Posted by sean92t0 View Post
    you're right, i read it wrong.
    So what is your answer? And how did you get it?
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  7. #7
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    well before that, consider this:

    3^668 * 3 * 1 < 3^668 * 2 * 2
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  8. #8
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    Quote Originally Posted by sean92t0 View Post
    well before that, consider this:

    3^668 * 3 * 1 < 3^668 * 2 * 2
    Hmm ya... I think I should try a new line of approach... I am trying it
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    Read this.

    The maximum product is indeed 2^2 \cdot 3^{668}.
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