A = 1 + 2 + 22 + ... + 22009 + 22010
= ( 1 + 2 + 22 ) + ( 23 + 24 + 25 ) + ... + ( 22008 + 22009 + 22010 )
= 7 + 23( 1 + 2 + 22 ) + ... + 22008( 1 + 2 + 22 )
= 7 + 23.7 + ... + 22008.7
= 7( 1 + 23 + ... + 22008 ) chia hết cho 7
hay A chia 7 dư 0
Ta có A = 1 + 2 + 22 + ... + 22009 + 22010
=> A - 3 = 22 + 23 + 24 + 25 + 26 + 27 + .... + 22008 + 22009 + 22010
= (22 + 23 + 24) + (25 + 26 + 27) + .... + (22008 + 22009 + 22010)
= 22(1 + 2 + 22) + 25(1 + 2 + 22) + ... + 22008(1 + 2 + 22)
= (1 + 2 + 22)(22 + 25 + ... + 22008)
= 7(22 + 25 + ... + 22008) \(⋮\)7
Vì \(A-3⋮7\)
=> A : 7 dư 3
Vậy A : 7 dư 3