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Message: <blockquote data-quote="Vyper" data-source="post: 1349465" data-attributes="member: 14683">Corran:The solutions that were posted by now are very good. But it's possible to solve the riddle completely without a computer or even a calculator. As you said, the main problem is the big solution space.Well, you can reduce it significantly.For example, every even number is the sum of two primes (Goldbach conjecture - this hasn't been proven, but it's certainly true for the small numbers we are talking about). So the sum must be odd.If one of the numbers is a prime number >50, B would immediately know the numbers. So none of the numbers can be a prime >50, and since A must know this, too, the sum cannot be 53+x with 2 <= x < 100, neither can it be 97+x with 2 <= x < 100. Thus, the sum must be less than 55. (197 and 198 can be eliminated easily.)So we are left with about 25 odd integers. We still need to find the ones that cannot be the sum of two primes. Every odd number is the sum of an odd and an even number. The only even prime number is 2, so the only remaining odd numbers that do not have the property P can be written as 2+p (with p prime).So, if I didn't make a mistake in my explanations, only 11 numbers are left now.btw, there is a complete (and quite elaborate) analytical solution of the problem that was posted to a German newsgroup (hence it's German): <a href="http://groups.google.de/groups?selm=9ualmd%247dcff%241%40ID-58404.news.dfncis.de" target="_blank">http://groups.google.de/groups?selm=9ualmd$7dcff$1@ID-58404.news.dfncis.de</a>However, I probably wouldn't use this puzzle in a real rpg, since it's much too difficult.</blockquote>

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