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Homework Help: A math prpblem I know the answer to - how do you do it?

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Find positive numbers n and a1,a2, ... , an such that a1 + ... + an = 1000 and the product a1a2...an is as large as possible.

    2. Relevant equations
    The sequence 2,3,4,5,6,7,8,9... to replace 1000

    3. The attempt at a solution

    The solution is 3^332 + 2^2, however, I want to understand how this solution was reached.
    The example given gives that sequence above to replace 1000, but I'm at a loss as to how to continue to this answer.
  2. jcsd
  3. Mar 23, 2009 #2


    Staff: Mentor

    How is this sequence relevant? How does a sequence replace 1000?
    Like you, I don't understand how this solution was reached. I also don't understand why 3^332 + 2^2 is a solution to this problem or even what this number represents. The problem asks you to find n and the numbers a1, a2, etc.

    3^332 is a huge number (about 2.54 X 10^158). I don't have a calculator that can add 4 to this number.

    If you look at a few values of n, you can find solutions. For example, if n = 2, then a1 = a2 = 500 gives you the right sum, and multiplies to the largest value.

    If n = 3, a1 = 333, a2 = 333, a3 = 334 produces a product of 37,036,926. I don't think you can get a larger product with three numbers whose sum is 1000.

    If n = 4, a1 = a2 = a3 = a4 = 250 gives the largest product.
  4. Mar 23, 2009 #3
    Well I can quickly realise that;

    1/2(n)(n+1) = SUM( a , ... , an] )


    1000 = 1/2 (n)(n+1) = SUM( a , ... , an )

    Expanding out;

    1/2 n2 + 1/2 n = 1000


    2000 = n2 + n

    But that's not going to help you much trying to find the products of two consequtive terms. Hence I would hazard we have a power series that converges to 1000.


    SUM( a , ... , an ) = SUM( anq)

    where q is some power. Looking at your answer, this looks good as you have a power in it. Therefore;

    1000 = SUM( anq) = anq + a(n+1)q + ... + a(n+k)q

    Now of course if you want the product of 'an' and 'a(n+1)' to be as great as possible then you don't want to be summing up to 'a(n+k)' Hence you just want the first two terms. Therefore we can write;

    1000 = anq + a(n+1)q

    Do a bit of jiggery pokery;

    1000 = a (nq + (n+1)q)

    ...and I'm guessing there is format to solve this someway but its escaping me right now. Hence I'm just going to plug the numbers in;

    1000 = a (1 + 2q)

    remember 1 to the power of anything is still 1, and that the product of the first two terms can be written;

    Big number = a*(a+2q) = a2 + a(2q)

    Now this looks like one of the equations above. Therefore I would hazard you only need to solve the simultanious equations;

    1000 = a (1 + 2q)
    Big number = a2 + a(2q)

    I'm thinking ratios myself....

    But perhaps that has given you some ideas to solve the question...? Sorry, as I don't know the particular standard method I can't describe this particular problem with a simmilar question, and I don't think I'm meant to post up full solutions. But this might help you.

  5. Mar 23, 2009 #4


    User Avatar
    Science Advisor

    It's easier to think about this problem if we're not restricted to integers.

    First, consider fixed n. We want to divide some constant c (=1000 in the problem) into n pieces such that their product is maximized. It's not too hard to show that the optimal thing to do is to divide c into n equal pieces, each with value c/n, so that the product is (c/n)^n.

    Now, what value of n maximizes this? It turns out to be n=c/e, so that each piece has the value c/n = c/(c/e) = e.

    If each piece must be an integer, the closest integer to e is 3. So we should divide c up into as many 3's as possible. For c=1000, this means 333 3's and one 1, whose product is 3^333 * 1.

    But we can do slightly better by trading in the 1 and one 3 for two 2's, so that we have instead 332 3's and two 2's, whose product is 3^332 * 2^2. This is larger by a factor of 4/3.

    After that, trading in 3's for 2's (or anything else) always lowers the product, so the optimal division is 332 3's and two 2's.

    I don't think this "proof" would satisfy a mathematician, though.
  6. Mar 23, 2009 #5
    I found the book in which this problem came from on Google Books (Problem Solving Through Problems), surprisingly,

    here's what the book has to say:

    i and ii are obvious, the reasons in parenthesis are the ones I don't understand (iii and iv)
  7. Mar 24, 2009 #6


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    Science Advisor

    (iii) means that one 4 is equivalent to two 2's; for example, in the case of 1000, there is a second solution, 332 3's and one 4; this gives the same sum (1000) and product as the original solution, 332 3's and two 2's. So we can discard 4's if we are only looking for one of the possible solutions.

    (iv) Suppose I claim to have a solution with three 2's. You can improve on it by replacing my three 2's with two 3's. The sum will be unchanged, since 2+2+2 = 3+3, but the product will go up, since 3x3 > 2x2x2. So my claim must be false.

    A cute problem.
    Last edited: Mar 24, 2009
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