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Topic: 271 (Read 616 times) |
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tereferekuku
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"Write 271 as the sum of positive real numbers so as to maximize their product."
SOLUTION:
e^99*e^(271/e-99)
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towr
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Re: 271
« Reply #1 on: Sep 21st, 2009, 1:29pm » |
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I don't see any + sign, or a  . So at the very least you haven't written 271 as the sum of anything.
And e^99 + e^(271/e-99) isn't 271 either; it's rather a long way off.
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| « Last Edit: Sep 21st, 2009, 1:30pm by towr » |
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tereferekuku
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Re: 271
« Reply #2 on: Sep 21st, 2009, 3:21pm » |
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OK, sum_1^99{e} + e^{271/e-99}. Is that better?
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Benny
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Re: 271
« Reply #3 on: Sep 21st, 2009, 3:35pm » |
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I worked on this problem just few days ago.
See Here
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towr
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Re: 271
« Reply #4 on: Sep 22nd, 2009, 1:45am » |
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on Sep 21st, 2009, 3:21pm, tereferekuku wrote:| OK, sum_1^99{e} + e^{271/e-99}. Is that better? |
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It's about 271.114269, though; not 271. But it's very close to the solution Ben linked to.
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tereferekuku
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Re: 271
« Reply #5 on: Sep 22nd, 2009, 3:21am » |
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Yes, of course, I meant sum_1^99{e} + e*{271/e-99}, which adds up just fine to 271, but the issue I forgot about is that ALL factors should be equal in order to maximize the product, hence Ben's solution is still slightly bigger than mine.
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R
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Re: 271
« Reply #6 on: Sep 23rd, 2009, 9:11pm » |
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What is so special about 271?
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Benny
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Re: 271
« Reply #7 on: Sep 23rd, 2009, 10:06pm » |
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Consider, for example, 271 is the smallest prime p such that (p - 1) and (p + 1) are each divisible by a cube greater than one.
http://primes.utm.edu/curios/page.php/271.html
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towr
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Re: 271
« Reply #8 on: Sep 24th, 2009, 12:21am » |
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on Sep 23rd, 2009, 9:11pm, R wrote:| What is so special about 271? |
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In this case? It's the second closest integer to 100*e
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R
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Re: 271
« Reply #9 on: Sep 24th, 2009, 12:57am » |
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on Sep 23rd, 2009, 10:06pm, BenVitale wrote:
lol 
Maths have a story behind every number. 
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