Prove: For all integers n, 21|(3n^7 + 7n^3 + 11n)

8 years agoI'm confused by your notation. Is the 21 part of the polynomial?

8 years agono 21 divides (3n^7 + 7n^3 + 11n)

8 years agoOhhh, I see. I suggest you use the back slash / next time. It's more clear that way. So for all integers n, what are you trying to prove? You didn't make that very clear either.

8 years agoProve that for all integers n, then 21 divides (3n^7 + 7n^3 + 11n) <-this number

8 years agoAre you trying to say that that polynomial is divisible by 21 for all integers n?

8 years agofor any integer n, then (3n^7 + 7n^3 + 11n) is divisible by 21 with no remainders

8 years agoit doesn't give you a whole lot of information to work from

8 years agoI'm actually not quite sure how to approach this. The polynomial is not divisible. Maybe a recursive equation or something. I'll think about it. i'll get back to you if I figure it out.

8 years agoI searched on google and found some other solutions, but I don't really understand them that well http://answers.yahoo.com/question/index?qid=20091102201309AA4bGAA

8 years agofor 4a iii) what did u call the proposition- in my notes it doesnt have a name

8 years agoI didn't call any proposition. I just checked it on a calculator

8 years agoo ok

8 years agoIs it the modular math that's throwing you off? mod x means remainder x. That proof looks good.

8 years agoyeah I was having trouble following the modular math, I've spent some time working thought it though and I think I've got an idea of where it's coming from. Congruences are new to me so it's still a little confusing. Thanks for looking into it for me though.

8 years agoi dont get how he divided by 3

8 years ago