Ta có:
`\qquad 2014 ≡ 1 (mod \ 11)`
`=>2014^{2013} ≡1^{2013} (mod \11)`
`=>2014^{2013} ≡ 1 (mod\ 11)` $(1)$
`\qquad 2015 ≡ 2 (mod\ 11)`
`=>2015^5 ≡ 2^5 (mod \11)`
`=>2015^5 ≡ 10 (mod \11)`
`=>(2015^5)^2≡10^2 (mod \11)`
`=>2015^{10} ≡ 1 (mod\ 11)`
`=> (2015^{10})^{201}≡ 1^{201} (mod \ 11)`
`=>2015^{2010} ≡1 (mod \11)` $(2)$
`\qquad 2015≡ 2 (mod\ 11)`
`=>2015^4≡2^4 (mod \11)`
`=>2015^4≡ 5 (mod\ 11)` $(3)$
Từ $(2);(3)$ suy ra:
`\quad 2015^4 .2015^{2010}≡ 5.1(mod\ 11)`
`=>2015^{2014} ≡ 5 (mod \11)` $(4)$
Từ $(1);(4)$ suy ra:
`2014^{2013}+ 2015^{2014}≡ 1+5=6(mod\ 11)`
Vậy `2014^{2013}+2015^{2014}` chia $11$ dư $6$
