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Title: 271 Post by rugga on Aug 28th, 2002, 4:30pm Quote:
I like this one because I initially didn't suspect what turned out to be the solution, but after some trial and error with smaller numbers I took a wild guess which turned out correct. Here's some hints at a proof (copying the style of a different post by S. Owen): Hint 1: The way to maximize the product of n numbers which sum to C is to make all n numbers equal. So the largest product is (C/n)n. So then the question becomes what value of n maximizes (C/n)n, where n is a positive integer. Hint 2: The derivative can find the real-valued n which maximizes the above expression. Then take the closest integer on either side and see which gives a larger answer. To take Dn(C/n)n, rewrite it as Dnen ln (C/n). Then apply the chain rule and a few steps to get (C/n)n (ln C - ln n - 1). Setting this equal to 0, you end up with ln n = ln C -1. Raising both sides to the power of e: n = Ce-1 = C/e Answer: For this problem C=271, so n~99.695. It's a fair bet 271 was chosen so that n would come out even, but check just to make sure: (271/99)99 ~ 1.977 x 1043 (271/100)100 ~ 1.981 x 1043 Anyone have a more intuitive explanation? I'm not sure I really understand why the math works out the way it does. |
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Title: Re: 271 Post by jk on Aug 29th, 2002, 8:38am I looked at recursively breaking 271 (or any number) down into addends (is that the word?), Arbitrarily, pick two numbers such that: N + M = 271 then, break down N and M: (N0 + N1) + (M0 + M1) = 271 Our product (N0 * N1)(M0 * M1), will grow as long as each component can be advantageously broken down. So, how far do you go? 1 is of course a loser: (N-1) * 1 < N 2*2 breaks even for 4 3*2 > 5 good 3*3 > 6 better Any addend larger than 4 can be broken down to advantage. I'd love something more elegant, but you can empirically plug in numbers and see that: For 6: 3*3 > 2*2*2 For 8: 3*3*2 > 2*2*2*2 Three seems to be the lowest advantageous breakdown, so I propose: 271 = 4+(89 * 3) 4*3^89 = 1.16... * 10^43 |
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Title: Re: 271 Post by AlexH on Aug 29th, 2002, 10:44am jk: the problem specified positive real numbers, not positive integers so rugga's solution is the correct one. rugga: In terms of finding the solution intuitive I'll say this ... note that maximizing (C/x)x is the same (substitute y=C/x) as maximizing yC/y = (y1/y)C. Since we're dealing with numbers > 1, this is the same thing as maximizing y1/y which is a standard early problem in derivatives. |
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Title: Re: 271 Post by jk on Aug 29th, 2002, 12:50pm OUCH! Yes, I misread the question. mea maxima culpa |
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Title: Re: 271 Post by rugga on Sep 1st, 2002, 2:44am Quote:
Thanks AlexH. After thinking about it some more here's something I came up with. The question is about maximizing the product of a bunch of numbers, which is the same as maximizing the sum of their logarithms. So we really want to find the number which has the largest log (or the most "multiplicative power") per unit: maximize (log x) / x Again using basic calculus, this is maximized when x = e (doesn't matter what the base of the log is). This helped me gain some insight. I'm not sure if it will resonate with others, but I hope someone might find it useful. BTW, I see that maximizing (log x) / x is the same as maximizing y1/y as you suggest. My formula is the log of your formula. |
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Title: Re: 271 Post by NickH on Sep 1st, 2002, 11:53am Another approach is to note that we seek the least integer x such that (271/x)^x > (271/(x+1))^(x+1). Simplifying, we seek the least x such that x*(1 + 1/x)^(x+1) > 271. We know that x is reasonably large, in which case (1 + 1/x)^(x+1) ~= e. So we seek the least x such that x > 271/e. This gives an intuitive feel as to why the terms (2.71, in this case) are close to e. It can be made more robust, if required. Nick |
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