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Wildan Bagus W

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On 11/7/2017 at 2:56 PM, Wildan Bagus W said:

Bisa pakai teleskopik

Dari hint:

Karena $m! \cdot m = (m+1)! - m!$ maka $S = (k+1)! -1$. Jadi $S \cong -1\mod (k+1)$. Akibatnya $S \cong k \mod k$.

nulis garis 3 (kongruensi modulo) gimana sih?

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8 hours ago, OnePunchMan said:

 

Dari hint:

Karena m!m=(m+1)!m!m!⋅m=(m+1)!−m! maka S=(k+1)!1S=(k+1)!−1 . Jadi S1mod(k+1)S≅−1mod(k+1) . Akibatnya SkmodkS≅kmodk .

nulis garis 3 (kongruensi modulo) gimana sih?

Itu pakai \equiv, jadnya $\equiv$

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