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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Tìm số dư khi chia cho các số sau (dùng đồng dư) : `a) (2014^{2015}+2015^{2016}):17` `b) (2014^{2013}+2015^{2014}):11`

Toán Lớp 8: Tìm số dư khi chia cho các số sau (dùng đồng dư) :
a) (2014^{2015}+2015^{2016}):17
b) (2014^{2013}+2015^{2014}):11

Comments ( 1 )

  1. Giải đáp:
    $a)\quad 16$
    $b)\quad 6$
    Lời giải và giải thích chi tiết:
    $a)\quad (2014^{2015} + 2015^{2016})\ :\ 17$
    Ta có:
    $+)\quad 2014\equiv 8\pmod{17}$
    $\Leftrightarrow 2014^{2015}\equiv 8^{2015}\pmod{17}$
    $\bullet\quad 8^8\equiv 1\pmod{17}$
    $\Leftrightarrow (8^8)^{251}\equiv 1\pmod{17}$
    $\Leftrightarrow 8^{2008}\equiv 1\pmod{17}$
    $\bullet\quad 8^7\equiv 15\pmod{17}$
    $\Rightarrow 8^{2008}.8^7\equiv 15 \pmod{17}$
    $\Rightarrow 8^{2015}\equiv 15\pmod{17}$
    $\Rightarrow 2014^{2015}\equiv 15\pmod{17}$
    $+)\quad 2015\equiv 9\pmod{17}$
    $\Leftrightarrow 2015^{2016}\equiv 9^{2016}\pmod{17}$
    $\bullet\quad 9^8\equiv 1\pmod{17}$
    $\Leftrightarrow (9^8)^{252}\equiv 1\pmod{17}$
    $\Leftrightarrow 9^{2016}\equiv 1\pmod{17}$
    $\Rightarrow 2015^{2016}\equiv  1\pmod{17}$
    Do đó:
    $(2014^{2015} + 2015^{2016})\equiv 16 \pmod{16}$
    Vậy $(2014^{2015} + 2015^{2016})\ :\ 17$ dư $16$
    $b)\quad (2014^{2013} + 2015^{2014})\ :\ 11$
    Ta có:
    $+)\quad 2014\equiv 1\pmod{11}$
    $\Leftrightarrow 2014^{2013}\equiv 1\pmod{11}$
    $+)\quad 2015\equiv 2\pmod{11}$
    $\Leftrightarrow 2015^{2014}\equiv 2^{2014}\pmod{11}$
    $\bullet\quad 2^{10}\equiv 1\pmod{11}$
    $\Leftrightarrow (2^{10})^{201}\equiv 1\pmod{11}$
    $\Leftrightarrow 2^{2010}\equiv 1\pmod{11}$
    $\bullet\quad 2^4\equiv 5\pmod{11}$
    $\Rightarrow 2^{2010}.2^4\equiv 5\pmod{11}$
    $\Rightarrow 2^{2014}\equiv 5\pmod{11}$
    $\Rightarrow 2015^{2014}\equiv 5\pmod{11}$
    Do đó:
    $(2014^{2013} + 2015^{2014})\equiv 6\pmod{11}$
    Vậy $(2014^{2013} + 2015^{2014})\ :\ 11$ dư $6$

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222-9+11+12:2*14+14 = ? ( )