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A puzzle

Posted By: Rich Munitz
Date: Friday, 5 November 2010, at 4:40 a.m.

In Response To: A puzzle (Sam Pottle)

"For any n>0, 2^(n-1) has n factors."

Duh! Must have slept through the first day of number theory class.

Interestingly, the product of any n different primes has 2^n-1 factors.

Messages In This Thread

• A puzzle
  Bob Koca -- Wednesday, 3 November 2010, at 4:20 p.m.
  • A puzzle
    storm -- Wednesday, 3 November 2010, at 4:51 p.m.
    • A puzzle
      Bob Koca -- Wednesday, 3 November 2010, at 5:27 p.m.
  • A puzzle
    tansley -- Thursday, 4 November 2010, at 5:17 p.m.
    • A puzzle
      storm -- Thursday, 4 November 2010, at 5:33 p.m.
    • A puzzle
      Bob Koca -- Thursday, 4 November 2010, at 7:43 p.m.
      • A puzzle
        Rich Munitz -- Thursday, 4 November 2010, at 9:18 p.m.
        • A puzzle
          Sam Pottle -- Thursday, 4 November 2010, at 9:51 p.m.
          • A puzzle
            Rich Munitz -- Friday, 5 November 2010, at 4:40 a.m.
            • A puzzle
              Bob Koca -- Friday, 5 November 2010, at 2:49 p.m.