A = 20 + 21 + 22 + 23 + ... + 22009 + 22010
=> A = 20 + ( 21 + 22 ) + ( 23 + 24 ) + ... + ( 22009 + 22010 )
=> A = 20 + 2 ( 1 + 2 ) + 23 ( 1 + 2 ) + ... + 22009 ( 1 + 2 )
=> A = 20 + 2 . 3 + 23 . 3 + ... + 22009 . 3
=> A = 1 + 3 ( 2 + 23 + ... + 22009 )
Vì : 3 ( 2 + 23 + ... + 22009 ) \(⋮\)3 => A chia cho 3 dư 1
Vậy : A chia cho 3 dư 1