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OpenStudy (ikram002p):
lolz zzr
OpenStudy (zzr0ck3r):
@Astrophysics said it was 2
OpenStudy (zzr0ck3r):
:P
OpenStudy (astrophysics):
That was a mistake! I thought we were multiplying at first >_>
OpenStudy (zzr0ck3r):
If you are an astrophysicists that gets 2 confused with 2^2013 we are all in trouble
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OpenStudy (zzr0ck3r):
lol jk
OpenStudy (astrophysics):
zz! lmao
OpenStudy (zzr0ck3r):
I thought it was 2 at first as well
OpenStudy (tanya123):
I'll be back with a more intermediate question.
OpenStudy (astrophysics):
So did nin, he was acting like he knew, but he gave me a medal right away.
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OpenStudy (camerondoherty):
i got it wrong :c
OpenStudy (astrophysics):
Nin doesn't know pellet
OpenStudy (camerondoherty):
o.o
OpenStudy (ikram002p):
ok @zzr0ck3r quick Qn for u ,
whats the remainder of 2^2013 when u divide it by 7 :P
OpenStudy (zzr0ck3r):
this was a good question, in all my years of doing math I don't think I have seen this(or a symmetrical) question.
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OpenStudy (zzr0ck3r):
find a power of 2 for which has a nice remainder
OpenStudy (zzr0ck3r):
@ikram002p what level number theory class? there are a few ways to do that. you can use fermats little theorem or the long way of finding nicer ways to write it....
OpenStudy (ikram002p):
well lol i randomly posted it for fun :P
dnt take it serously hehe