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22850Re: [PrimeNumbers] Post-Cartesian Puzzle

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  • Mathieu Therrien
    Jul 14, 2011
      Thx, I will rework it.

      From: Tom Hadley <kctom99@...>
      To: Mathieu Therrien <mathieu344@...>
      Cc: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>
      Sent: Thursday, July 14, 2011 4:41:26 PM
      Subject: Re: [PrimeNumbers] Post-Cartesian Puzzle

       Mathieu Therrien <mathieu344@...> wrote:

      >If I understood the purpose of Sigma(N/M) = N/P_1 = M*P_2   for P = P_1 * P_2 correctly,
      >then Many solution are possibles as long as (M+1) is divided by 2 only once 
      >for example u have m=5 ; N = P_2 * m * 3 = 15*P_2  and as long that P_2 is odd
      >So m=5 and N=45 is 1 solution
      I think you have misunderstood the sigma() function.  In Pari-GP, sigma(x) is the sum of the divisors of x.  So sigma(9) = 1+3+9 = 13.
      The puzzle is: Find a pair of odd integers (N,m) with m|N,
      sigma(N/m)*(1+m) = 2*N, and bigomega(m) = 1.
      The proposed solution, N=45, m=5 doesn't work, since
      sigma(45/5)*(1+5) = sigma(9)*6 = 13*6 = 78, which is not 2*N=2*45=90.
      Tom Hadley

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