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anonymous
 one year ago
@ganeshie8
anonymous
 one year ago
@ganeshie8

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok. Let me open teamview

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0uhm... can not connect? O.o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it the same ID you game me last time?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1nope that was finished.. you need to start a fresh meeting.. click on Meeting, then you will see "presentation" icon on left pane

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 uhm... hello? :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1It says you haven't started the presentation...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0O.o how do I start it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let : \[f(x) = \sum\limits_{k=0}^{n}a_kx^k\] \[g(x) = \sum\limits_{k=0}^{n}a_k(x+jp)^k\] show that \(f(x)\equiv g(x) \pmod{p}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\((x+jp)^k = x^k + p*(some integer)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me skip typing the sum, so it's x^i (jp)^(ki), i from 0 to k ? oh, x^0 (jp)^(k0) + x^1 (jp)^(k1) + x^2 (jp)^(k2) + ... + x^(k1)(jp)^1 + x^k (jp)^0 yeah I see p as a factor of every term except the last term p*(some integer) + x^k yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\((x+jp)^k = x^k + p*(some integer) \equiv x^k + 0 \pmod{p}\)
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