The last digit of 5^n for n>0 has to be 5. 1 = 5^0 5 = 5^1
If 5^k is ...5, that is (10*m)+5, then 5^k+1 5*5^k 5*(10*m)+5 (10*5*m)+25 (10*2+5*m)+5 Similar proof for the last 2 digits of 5^n for n>1. ----- Original Message ----- From: Ian Gorse <[EMAIL PROTECTED]> Date: Friday, September 26, 2008 14:14 Subject: Re: [Jprogramming] Difference between 5^100x and x: 5^100 To: Programming forum <[email protected]> > > > > As a check, you know that the last digit of 5^n for n>0 has > > to be 5, so the x:5^100 result can not be the exact result. > > (The last 2 digits of 5^n for n>1 have to be 25, etc.) > > > > > Thanks for the reply Roger, but I have a very simple question > regarding the quoted statement. > > Why? > > Please feel free to reply a link or search term if its out of > scope of > this conversation. ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
