chọn B. 27/1000

x=36

y=   1

       9

Ta có:  \(5^3\equiv1\left(mod31\right)\)

=> \(\left(5^3\right)^{671}\equiv1\left(mod31\right)\)

=> \(\begin{cases}\left(5^3\right)^{671}\cdot5^2\equiv25\left(mod31\right)\equiv25\left(mod31\right)\\\left(5^3\right)^{671}\cdot5^3\equiv5^3\left(mod31\right)\equiv1\left(mod31\right)\\\left(5^3\right)^{671}\cdot5^3\cdot5\equiv5^4\left(mod31\right)\equiv5\left(mod31\right)\end{cases}\)

=> \(\begin{cases}5^{2015}\equiv25\left(mod31\right)\\5^{2016}\equiv1\left(mod31\right)\\5^{2017}\equiv5\left(mod31\right)\end{cases}\)

=> \(5^{2015}+5^{2016}+5^{2017}\equiv25+5+1\left(mod31\right)\equiv0\left(mod31\right)\)

Vậy \(5^{2015}+5^{2016}+5^{2017}⋮31\left(đpcm\right)\)

Đáp số : 98