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11:20
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A: Remainder of division.

lab bhattacharjee$$2014^{2013}\equiv14^{2013}\pmod{20}$$ and Carmicheal Function $\displaystyle\lambda(20)=4,2013\equiv3\pmod4$ $$\implies14^{2013}\equiv14^3\pmod{20}$$ Again,$$14^3=(10+4)^3=10^3+\binom3110^2\cdot4+\binom3210^1\cdot4^2+4^3\equiv4\pmod{20}$$ $$\implies2014^{2013}=20a+4$$ for some integer $a$ ...

Artemisia
Artemisia
Thank you :) Why have you chosen mod 20?
lab bhattacharjee
lab bhattacharjee
@Artemisia, You have hinted $$2^{10}\equiv-1\pmod{41}\implies2^{20}=(2^{10})^2\equiv?$$
Artemisia
Artemisia
OH! Ok I understand now :) Thank you! :)
lab bhattacharjee
lab bhattacharjee
@Artemisia, My Pleasure. See also: math.stackexchange.com/questions/905906/…
Artemisia
Artemisia
I have a question. What is $a$? I mean, shouldn't that be determined as well?
lab bhattacharjee
lab bhattacharjee
11:20
@Artemisia, Thta's why I reached at $20$. $a$ is an integer, right and $1^a=?$
Artemisia
Artemisia
Oh I see! So it's $16 (mod 41)$? I mean, that's the answer?
Hello :)
so I am not sure if I understand this correctly (if at all)
lab bhattacharjee
lab bhattacharjee
hi
Please pinpoint your confusion
Artemisia
Artemisia
I understood your simplification... but is there a way to do this using the Chinese Remainder theorem?
lab bhattacharjee
lab bhattacharjee
Yes
Artemisia
Artemisia
Because this was given as an exercise to that and I thought that's the way to go about it
lab bhattacharjee
lab bhattacharjee
11:24
You can find 2014^(2013)mod(20) by find 2014^(2013)mod(4) and 2014^(2013)mod(5)
Artemisia
Artemisia
hmmm ok...
and once that is done then I will get 2014^(something)?
lab bhattacharjee
lab bhattacharjee
no
Artemisia
Artemisia
2013 mod 4 is 1 mod 4 ... and 2013 mod 5 is 3 mod 5...
lab bhattacharjee
lab bhattacharjee
2014^n ≡0 mod 4 for integer n>0 and 2013≡3(mod 5) 2014^(2013)2014^(3)≡(-1)^3
sorry n>1
Artemisia
Artemisia
ok...
ok let me try. 2013(mod 20) can be split using the Chinese remainder theorem
to mod 4 and mod 5
so then I need to find 2013 mod 4 and 2013 mod 5, no?
and "multiply" the result?
oh wait you have solved it with the 2014 as well
lab bhattacharjee
lab bhattacharjee
11:33
not multiply
Chinese Remainder
Artemisia
Artemisia
oh ok ok
let me do it again
so this way I get 16 (mod 41).
sorry, -16
lab bhattacharjee
lab bhattacharjee
why -16
Artemisia
Artemisia
because 2^10=-1(mod 41)...?
lab bhattacharjee
lab bhattacharjee
But we have (-1)^(2a)
Artemisia
Artemisia
oh then +1 :) sorry ... I got pretty stressed out seeing this question. I never completely understood modular arithmetic, to be honest.
lab bhattacharjee
lab bhattacharjee
11:47
understood?
Artemisia
Artemisia
2014^n ≡0 mod 4 n>1 and 2013≡3(mod 5), 2014^(2013)2014^(3)≡(-1)^3 . This step is tripping me over.
lab bhattacharjee
lab bhattacharjee
Please go through mathworld.wolfram.com/Congruence.html esp. the properties
Artemisia
Artemisia
oh ok! I get it now :)
Thank you so much for your time! :)
lab bhattacharjee
lab bhattacharjee
welcome

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