tự túc là hạnh phúc

$\textbf{1.}$

Ta có $5^{2017}+5^{2015}=5^{2015}(5^2+1)$

$=5^{2015}\cdot26$

$=5^{2015}\cdot13\cdot2.$

Vì $13\mid13\cdot2$ nên $13\mid\left(5^{2017}+5^{2015}\right).$

Vậy $5^{2017}+5^{2015}$ chia hết cho $13.$

$\textbf{2.}$

Giả sử $a^{2014}+b^{2015}+c^{2016}\vdots6.$

Ta có $a^{2016}-a^{2014}=a^{2014}(a^2-1)$

$=a^{2014}(a-1)(a+1).$

Vì $a(a-1)(a+1)\vdots6$ nên $a^{2014}(a-1)(a+1)=a^{2013}\cdot a(a-1)(a+1)\vdots6.$

Suy ra $a^{2016}\equiv a^{2014}\pmod6.$

Tương tự, $b^{2017}-b^{2015}=b^{2015}(b^2-1)$

$=b^{2014}\cdot b(b-1)(b+1)\vdots6,$ nên $b^{2017}\equiv b^{2015}\pmod6.$

Lại có $c^{2018}-c^{2016}=c^{2016}(c^2-1)$

$=c^{2015}\cdot c(c-1)(c+1)\vdots6,$ nên $c^{2018}\equiv c^{2016}\pmod6.$

Cộng ba đồng dư trên, $a^{2016}+b^{2017}+c^{2018}\equiva^{2014}+b^{2015}+c^{2016}\equiv0\pmod6.$

Vậy $a^{2016}+b^{2017}+c^{2018}$ chia hết cho $6.$

Chứng minh

a) \(2\equiv-1\left(mod3\right)\)

\(\Rightarrow2^{1000}\equiv\left(-1\right)^{1000}\equiv1\left(mod3\right)\Rightarrow2^{1000}-1\equiv0\left(mod3\right)\Rightarrowđpcm\)

b) \(19\equiv-1\left(mod20\right)\)

\(\Rightarrow19^{45}\equiv\left(-1\right)^{45}\equiv1\left(mod20\right);19^{30}\equiv\left(-1\right)^{30}\equiv1\left(mod20\right)\)

\(\Rightarrow19^{45}+19^{30}\equiv0\left(mod20\right)\Rightarrowđpcm\)

dùng đồng dư là ra mà

thì cũng giống như 4 mũ 5 + 6 mũ 5 chia hết cho 5

Ta có \(2015^{2015}-2015^{2014}=2015^{2014}.2015-2015^{2014}=2015^{2014}.\left(2015-1\right)=2015^{2014}.2014\) chia hết cho 2014 (đpcm).