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Artemisia

 Discussion between lab bhattacharjee …

Imported from a comment discussion on math.stackexchange.com/q...
Artemisia
Aug 23, 2014 11:51
Thank you so much for your time! :)
Artemisia
Aug 23, 2014 11:51
oh ok! I get it now :)
Artemisia
Aug 23, 2014 11:48
2014^n ≡0 mod 4 n>1 and 2013≡3(mod 5), 2014^(2013)2014^(3)≡(-1)^3 . This step is tripping me over.
Artemisia
Aug 23, 2014 11:45
oh then +1 :) sorry ... I got pretty stressed out seeing this question. I never completely understood modular arithmetic, to be honest.
Artemisia
Aug 23, 2014 11:42
because 2^10=-1(mod 41)...?
Artemisia
Aug 23, 2014 11:37
sorry, -16
Artemisia
Aug 23, 2014 11:34
so this way I get 16 (mod 41).
Artemisia
Aug 23, 2014 11:33
let me do it again
Artemisia
Aug 23, 2014 11:33
oh ok ok
Artemisia
Aug 23, 2014 11:31
oh wait you have solved it with the 2014 as well
Artemisia
Aug 23, 2014 11:30
and "multiply" the result?
Artemisia
Aug 23, 2014 11:30
so then I need to find 2013 mod 4 and 2013 mod 5, no?
Artemisia
Aug 23, 2014 11:29
to mod 4 and mod 5
Artemisia
Aug 23, 2014 11:29
ok let me try. 2013(mod 20) can be split using the Chinese remainder theorem
Artemisia
Aug 23, 2014 11:29
ok...
Artemisia
Aug 23, 2014 11:27
2013 mod 4 is 1 mod 4 ... and 2013 mod 5 is 3 mod 5...
Artemisia
Aug 23, 2014 11:25
and once that is done then I will get 2014^(something)?
Artemisia
Aug 23, 2014 11:25
hmmm ok...
Artemisia
Aug 23, 2014 11:24
Because this was given as an exercise to that and I thought that's the way to go about it
Artemisia
Aug 23, 2014 11:23
I understood your simplification... but is there a way to do this using the Chinese Remainder theorem?
Artemisia
Aug 23, 2014 11:22
so I am not sure if I understand this correctly (if at all)
Artemisia
Aug 23, 2014 11:22
Hello :)
Artemisia
Aug 23, 2014 11:20
Oh I see! So it's $16 (mod 41)$? I mean, that's the answer?
Artemisia
Aug 23, 2014 11:20
I have a question. What is $a$? I mean, shouldn't that be determined as well?
Artemisia
Aug 23, 2014 11:20
OH! Ok I understand now :) Thank you! :)
Artemisia
Aug 23, 2014 11:20
Thank you :) Why have you chosen mod 20?
 

 deutschsprachiger Raum

General discussion for german.stackexchange.com. You may speak...
Artemisia
Jan 10, 2014 16:21
It was really bothering me :)
Artemisia
Jan 10, 2014 16:21
Danke schoen :)
Artemisia
Jan 10, 2014 14:18
The song starts at 1:13
Artemisia
Jan 10, 2014 14:18
►
Artemisia
Jan 10, 2014 14:17
Hallo! I came across a song in an Indian sit-com that I watch and I suspect it's in German. Can anyone help me find this song?
 

 Chez Cosette

Discussion pour french.stackexchange.com. Bienvenue à tous ! Y...
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Artemisia
Jan 6, 2014 15:06
Merci beaucoup :)
Artemisia
Jan 6, 2014 15:05
maintenant, alors
Artemisia
Jan 6, 2014 15:05
Oh!
Artemisia
Jan 6, 2014 15:05
I have a question which was advised to me to be pitched in here
Artemisia
Jan 6, 2014 15:04
Allo!
 

 Mathematics

Associated with Math.SE; for both general discussion & math qu...
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Artemisia
Nov 23, 2013 06:43
cheers
Artemisia
Nov 23, 2013 06:43
haha sure :)
Artemisia
Nov 23, 2013 06:40
Yes
Artemisia
Nov 23, 2013 06:38
Thank you :)
Artemisia
Nov 23, 2013 06:38
Ok :)
Artemisia
Nov 23, 2013 06:37
Ah. Does every group have {e} subgroup?
Artemisia
Nov 23, 2013 06:33
Also, what are non-trivial subgroups? Subgroups without zero?
Artemisia
Nov 23, 2013 06:31
I need to show that two reflections on the complex plane are a rotation if the lines of reflection are intersecting, else they form a translation
Artemisia
Nov 23, 2013 06:31
I understand what this question means, but I don't know how to prove it
Artemisia
Nov 23, 2013 06:30
Hello :) I have a question on group theory
 

 Could you help me see my mistake?

Artemisia
Nov 22, 2013 15:17
Thank you :) Good luck to you too :)
Artemisia
Nov 22, 2013 15:17
I shall get back to frieze groups :)
Artemisia
Nov 22, 2013 15:17
:) Once again, thank you so much for all your help :)
Artemisia
Nov 22, 2013 15:16
I truly enjoy Physics :)