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 2003-06-30, 13:33 #1 xilman Bamboozled!     "๐บ๐๐ท๐ท๐ญ" May 2003 Down not across 3×73×11 Posts Primes and relatively primes. This challenge is in two parts, the first quite easy and the second harder. 12 has the property that all natural numbers less than it and relatively prime to it are themselves prime. a) What is the largest integer with this property? b) Why? Paul
 2003-06-30, 16:03 #2 hyh1048576   Jun 2003 26 Posts 30 is the largest. First,if p^2<S(the largest integer with this property),then p|S.(Otherwise p^2 is relatively prime to S but not prime).So if 7|S,we get 2|S,then 3|S,5|S,...... It's easy to prove that p1*p2*p3*...*pn>pn+1^2,where pn is the nth prime and n>3.This means 7|S will lead to "All the primes divides S", which is impossible!!! So 7 doesn't devide S,then S<49.JUST VERIFY IT :)
2003-06-30, 16:31   #3
xilman
Bamboozled!

"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across

3×73×11 Posts
Primes and relatively primes: proposed answer.

Quote:
 Originally Posted by hyh1048576 It's easy to prove that p1*p2*p3*...*pn>pn+1^2,where pn is the nth prime and n>3.

If it's easy to prove, then you should prove it. Until you have provided a valid proof, I don't think you have fully answered the question.

Of course, if you can prove it's false then you certainly haven't answered the question!

Paul

2003-07-01, 03:47   #4
hyh1048576

Jun 2003

26 Posts
Re: Primes and relatively primes: proposed answer.

Quote:
 Originally Posted by xilman If it's easy to prove, then you should prove it. Until you have provided a valid proof, I don't think you have fully answered the question. Paul
Really easy:
Using Chebyshev's theorem, pn+1&lt;2pn.So p1*p2*p3*...*pn>2*5*pn-1*pn>5*pn^2>4pn^2>pn+1^2,when n>4.When n=4,2*3*5*7>11^2. :)
It can also be proved without Chebyshev's theorem,but it's a little messy ;)

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