A = 11 + 112 + 113 + 114 + ... + 112013 + 112014
= (11 + 112) + (113 + 114) + ... + (112013 + 112014)
= 11(1 + 11) + 113(1 + 11) + .... + 112013(1 + 11)
= (1 + 11)(11 + 113 + ... + 112013)
= 12(11 + 113 + ... + 112013)
=> A \(⋮\)12 (ĐPCM)
A = 11 + 112 + 113 + .....+ 112014
A = (11 + 112) + (113 + 114) +...+ (112013 + 112014)
A = 11(1 + 11) + 113(1 + 11) +...+ 112013(1 + 11)
A = (1 + 11)(11 + 113 +...+ 112013)
A = 12 (11 + 113 +...+ 112013) \(⋮\)12 (vì 12 \(⋮\)12)
Vậy A \(⋮\)12