| Tired of seeing ads? Click here to upgrade to Elite Membership! |
Mathematics Forum
|
| Author | Message / Information |
| Euler Quote | Reply | | Challenge again... posted on: 6/14/2006 10:55:28 PM I am not nearly as well versed in number theory topics as I am in others, so this may be a pretty simplistic challenge, but nevertheless here it is, and if someone does solve, please feel free to post a new challenge of your own. Here it is: Solve for a: 4a+4=3(mod 6), (note: the equals sign should be replaced with the congruence symbol) Have fun! |
|
illoyd
Quote | Reply | |
Challenge again...
replied on: 6/15/2006 3:35:23 AM You put it back in its literal form: 4a+4= (n*6+3)/6 where n= 0,1,2,3 etc. solving a = (n*2-7)/8 |
|
Euler
Quote | Reply | |
Challenge again...
replied on: 6/15/2006 9:34:58 AM Sorry, but you'll have to try again! |
|
The Riemann Hypothesis
Quote | Reply | |
Challenge again...
replied on: 6/17/2006 7:36:26 AM 4a+4 = 3 + n*6 4a = 6n - 1 a = (6n - 1)/4 ?? |
|
Euler
Quote | Reply | |
Challenge again...
replied on: 6/17/2006 10:33:29 PM I would say that those question marks at the end of your post are close enough, since there is, in fact, no solution. Here is another, and this one does have a solution- Given a and b are different integers, p and q are different primes, aq(mod p)=1, and bp(mod q)=1, then prove: a/p + b/q > 1 Enjoy!! |
|
LinkBot
|
Gamers Wanted is looking for people to write game reviews and post news, |
|
|
| Tired of seeing ads? Click here to upgrade to Elite Membership! |
ChatArea.com Help & News Forums | Terms of Use | Contact ChatArea.com | Advertising
Powered By ChatArea.com - Get your free Society today! © Copyright 2003 Wewp!