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Re: don't ever give up! [Puzzle127]

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  • djbroadhurst
    ... Here is a far smaller (and far less elegant) solution. ... Solution: I use all 30 odd primes up to 127 and encode Q by the vector of their exponents:
    Message 1 of 10 , Aug 27, 2011
      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <d.broadhurst@...> wrote:

      > > Q = 7^((3^266-1)/2) * 3^133
      > Congratulations to William for an elegant
      > (if somewhat large:-) answer.

      Here is a far smaller (and far less elegant) solution.

      > Puzzle 127: Find an odd 127-smooth integer Q such that
      > gcd(Q^2,sigma(Q^2)) has 127 decimal digits.

      Solution: I use all 30 odd primes up to 127
      and encode Q by the vector of their exponents:

      {N=[82, 22, 13, 10, 19, 31, 7, 4, 1, 12, 4, 7, 4, 10,
      4, 7, 4, 4, 2, 13, 4, 10, 6, 6, 10, 10, 4, 3, 1, 4];
      Q=prod(k=1,30,prime(k+1)^N[k]);G=gcd(Q^2,sigma(Q^2));
      if(#Str(G)==127,print("valid solution"));}

      valid solution

      Comment: There are many similar solutions. For example
      we may divide Q by 13^9 and then multiply by any
      of these 12 prime powers:

      {M=[11^21, 19^30, 29^24, 37^18, 53^30, 61^27,
      67^12, 71^25, 97^21, 103^12, 109^7, 113^6];t=0;
      for(j=1,12,Q2=(Q/13^9*M[j])^2;G=gcd(Q2,sigma(Q2));
      if(#Str(G)==127,t++));print(t" more solutions");}

      12 more solutions

      David Broadhurst
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