Guess it should have been a problem involving a 3 digit number with 0 for the middle digit. Then, a base of -17 would be equivalent to a base of +17.
Regards, Richard Schuh -----Original Message----- From: The IBM z/VM Operating System [mailto:[EMAIL PROTECTED] On Behalf Of Brian Nielsen Sent: Wednesday, March 07, 2007 9:01 AM To: [email protected] Subject: Re: DASD cylinders On Tue, 6 Mar 2007 15:57:31 -0600, Adam Thornton <[EMAIL PROTECTED]> wrote: >I'll give you a dollar if you can show me a base in which 54 is more >than two and a half times as big as 27. Converting to mathematical terms: j=54 k=27 n=2.5 You want j/k > n, where j and k are in base b. Expressing j and k in terms of base b gives: j=54 base b = 5*b^1 + 4*b^0 = 5b+4 k=27 base b = 2*b^1 + 7*b^0 = 2b+7 Substituting into j/k > n: (5b+4) / (2b+7) > n, where n>2.5 Now choose an n > 2.5, and solve the following equation for b: (5b+4) / (2b+7) = n When n=3, (5b+4) / (2b+7) = 3 5b+4 = 3(2b+7) 5b+4 = 6b+21 b=-17 To double check, we substitute b=-17 into: (5b+4) / (2b+7) > 2.5, giving: (-85+4) / (-34+7) > 2.5 (-81)/(-27) > 2.5 3 > 2.5, which is indeed true. For larger values of n, b asymptotically approaches -3.5. For values of n=2.5+x, where x>0 and as x->0, b->negative infinity QED. A long time ago I read a mathematical paper on some interesting propertie= s of negative base number systems. Now it's earned me a dollar. Brian Nielsen
