Search found 6 matches

by Swaku
Fri Jan 29, 2010 4:05 am UTC
Forum: Mathematics
Topic: I am terrible at math.
Replies: 6
Views: 931

Re: I am terrible at math.

Hi My 2 bits worth. When I am doing badly for a particular subject, I am reminded of what my lecturer said. "Better make sure you KNOW what you DON'T KNOW" if you do then you can start to improve. Perhaps you can post of of the topics you are bad in then we can see perhaps how to present i...
by Swaku
Fri Jan 29, 2010 3:57 am UTC
Forum: Mathematics
Topic: I'm bad at maths. Any ideas why?
Replies: 36
Views: 4570

Re: I'm bad at maths. Any ideas why?

Well, I don't plan on my career to be maths-oriented, per se (law degree, hopefully), but I do see mathematics, particularly physics, as a potential hobby-like inspiration. I especially like the comment about facing mathematics not as mathematics but meditation; that was quite ingenious. I just hav...
by Swaku
Fri Jan 29, 2010 3:52 am UTC
Forum: Mathematics
Topic: Help with Combinatoric Homework
Replies: 8
Views: 1203

Re: Help with Combinatoric Homework

Thanks for the Compliment! Anyway for the tutorial yesterday my tutor had a more elegant solution so ill post it here: Qn : Show that there is a positive integer m such that 44^m - 1 is divisible by 7 Ans: We want to find out if there is an m such that (44^m - 1) = 0 mod 7 Consider 44^1 -1, ......, ...
by Swaku
Tue Jan 26, 2010 5:14 pm UTC
Forum: Mathematics
Topic: Help with Combinatoric Homework
Replies: 8
Views: 1203

Re: Help with Combinatoric Homework

So now the problem is reduced to "for what values of A is 2^A is a multiple is 8. the answer is when A is a multiple of 3 since 2^3A is always divisible by 8 and therefore 2^3A - 1 is divisble by 7. This line is wrong. For example when A=5, 2^A=32 which most definitely divides by 8, but 32-1=3...
by Swaku
Tue Jan 26, 2010 4:55 pm UTC
Forum: Mathematics
Topic: Help with Combinatoric Homework
Replies: 8
Views: 1203

Re: Help with Combinatoric Homework

I think this is it!!! rewrite (44^m - 1) as (42 + 2)^m - 1. Consider (42 + 2)^m. By binomial theorem we see that all the terms except the last are divisible by 7 because they have a "42" embedded in them eg for m=3 ,(42 + 2)^3 = 3c0 . 42^3 . 2^0 + 3c1 . 42^2 . 2^1 + .... + 3c3 . (42^0) . 2...
by Swaku
Tue Jan 26, 2010 1:58 pm UTC
Forum: Mathematics
Topic: Help with Combinatoric Homework
Replies: 8
Views: 1203

Help with Combinatoric Homework

Hey everyone, this is my first post to the xkcd forum. Im a 2nd year major in Applied Mathematics and I'm taking a class on Combinatorics ( My all time favorite topic) I am in need of a homework problem. I don't want any answers, just some helpful hints. Question: Show that there is a positive integ...

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