Prove: For all integers n, 21|(3n^7 + 7n^3 + 11n)
I'm confused by your notation. Is the 21 part of the polynomial?
no 21 divides (3n^7 + 7n^3 + 11n)
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.
Prove that for all integers n, then 21 divides (3n^7 + 7n^3 + 11n) <-this number
Are you trying to say that that polynomial is divisible by 21 for all integers n?
for any integer n, then (3n^7 + 7n^3 + 11n) is divisible by 21 with no remainders
it doesn't give you a whole lot of information to work from
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.
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
for 4a iii) what did u call the proposition- in my notes it doesnt have a name
I didn't call any proposition. I just checked it on a calculator
Is it the modular math that's throwing you off? mod x means remainder x. That proof looks good.
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.
i dont get how he divided by 3
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