flyers Posted April 16, 2002 Share Posted April 16, 2002 Originally posted by yashiro hmmm... They all ate the same (8/3)loaf or 2.666r A gave away (7/3)loaf or 2.333r B gave away (1/3)loaf or 0.333r C paid 8g for (8/3)loaf so 1g = (1/3)loaf Hence: A should get 7g and B gets 1g. Thankyou. :) Who said they all ate the same amount? In fact, the 1st 2 guys started eating before inviting the 3rd -- according to the 1st post. Link to comment Share on other sites More sharing options...
arsekicker Posted April 16, 2002 Share Posted April 16, 2002 "MAN B said that they should share the gold (4 pieces each) because everyone ate the same amount." MAN B said it Dufus. 7 to MAN A and 1 to MAN B is fair. MAN A gave up (15/3) - (8/3) = 7/3 2 2/3 (8/3) is the amount he ate. MAN B gave up (9/3) - (8/3) = 1/3 2 2/3 (8/3) is the amount he ate. 7/3 - Man A's contribution 1/3 - Man B's contribution. Me thinks. Link to comment Share on other sites More sharing options...
Miran Posted April 16, 2002 Share Posted April 16, 2002 I have a problem maybe someone can solve it: Let A be a ring with identity. Show that A is an integral domain if and only if deg(p(x)*q(x)) = deg(p(x)) + deg(q(x)) for all p(x), q(x) element of A[x]. Thanx. Link to comment Share on other sites More sharing options...
flyers Posted April 16, 2002 Share Posted April 16, 2002 Originally posted by arsekicker "MAN B said that they should share the gold (4 pieces each) because everyone ate the same amount." MAN B said it Dufus. 7 to MAN A and 1 to MAN B is fair. MAN A gave up (15/3) - (8/3) = 7/3 2 2/3 (8/3) is the amount he ate. MAN B gave up (9/3) - (8/3) = 1/3 2 2/3 (8/3) is the amount he ate. 7/3 - Man A's contribution 1/3 - Man B's contribution. Me thinks. Missed that, it's late. Dufus is a bit harsh, eh???:D Link to comment Share on other sites More sharing options...
CloudEngineer Posted April 16, 2002 Share Posted April 16, 2002 Originally posted by Knoxca I have a problem maybe someone can solve it: Let A be a ring with identity. Show that A is an integral domain if and only if deg(p(x)*q(x)) = deg(p(x)) + deg(q(x)) for all p(x), q(x) element of A[x]. Thanx. Say wat :s Link to comment Share on other sites More sharing options...
Quboid Posted April 16, 2002 Share Posted April 16, 2002 Ahhhh. I get it now. Yeah, that seems right. Brian is even more of a git than I first thought. Thanks. Knoxca - Nice try. Now do your homework yourself. :p Link to comment Share on other sites More sharing options...
EidrahS Posted April 16, 2002 Author Share Posted April 16, 2002 hehe this was not for homework.....Im doing ISLAMIC HIGHER EDUCATION studies and this too place many years ago. Anyway anotherway of looking at it: 8 pieces of bread, each man ate the same amount. So then beak each loaf into 3 pieces (hence for 3 men). You now have 24 pieces. From MAN A's loaf he ate 8 pieces and 7 of his pieces went to MAN C. From MAN B's loaf, he ate 8 pieces and 1 piece went to MAN C. So MAN A contributed 7 pieces while MAN B only contributed 1 piece to MAN C.............FAIR thats why u have 7 pieces of gold for MAN A and 1 for MAN B. We like to think with our hearts before our brain. hehehe.,..let the heart pump the blood folks.....and start using our brains a little bit. Link to comment Share on other sites More sharing options...
Wickedkitten Veteran Posted April 16, 2002 Veteran Share Posted April 16, 2002 well ok, i think that considering the fact that man A and man B bought the bread (obviously for 1 piece of gold each since C gave them 8 pieces of gold) Man A paid 5 for his, Man B paid 3 for his, Man C ate the bread as well but he paid for it so they should get what they originally paid in the first place. Man A gets 5 Man B gets 3 Link to comment Share on other sites More sharing options...
KnifeFace Posted April 16, 2002 Share Posted April 16, 2002 Originally posted by Kareemah hehe this was not for homework.....Im doing ISLAMIC HIGHER EDUCATION studies and this too place many years ago. Anyway anotherway of looking at it: 8 pieces of bread, each man ate the same amount. So then beak each loaf into 3 pieces (hence for 3 men). You now have 24 pieces. From MAN A's loaf he ate 8 pieces and 7 of his pieces went to MAN C. From MAN B's loaf, he ate 8 pieces and 1 piece went to MAN C. So MAN A contributed 7 pieces while MAN B only contributed 1 piece to MAN C.............FAIR thats why u have 7 pieces of gold for MAN A and 1 for MAN B. We like to think with our hearts before our brain. hehehe.,..let the heart pump the blood folks.....and start using our brains a little bit. yeah, but why did A give loaf to B? and B to A? doesn't make sence Link to comment Share on other sites More sharing options...
eth3l Posted April 16, 2002 Share Posted April 16, 2002 No gold pieces are rightfully Man B's there are no damages available here. There was no contract, and each person gave their bread free from any compensation. The gold pieces were a donation, and thus should follow the intent of the donor. As such each person should receive 4 pieces. Link to comment Share on other sites More sharing options...
nemo Posted April 16, 2002 Share Posted April 16, 2002 i should get all the gold. who cares about who brought the bread? Link to comment Share on other sites More sharing options...
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