Mathematics 34 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

I'm confused by your notation. Is the 21 part of the polynomial?

OpenStudy (anonymous):

no 21 divides (3n^7 + 7n^3 + 11n)

OpenStudy (anonymous):

Ohhh, 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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Are you trying to say that that polynomial is divisible by 21 for all integers n?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

it doesn't give you a whole lot of information to work from

OpenStudy (anonymous):

I'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.

OpenStudy (anonymous):

I 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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

o ok

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yeah 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.

OpenStudy (anonymous):

i dont get how he divided by 3

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