Ta có 2002 ⋮ 11 => 2004 - 2 ⋮ 11 => 2004 ≡ 2 (mod 11)
=> 20042004 ≡ 22004 (mod 11) , mà 210 ≡ 1 (mod 11) (vì 1024 - 1 ⋮ 11)
=> 20042004 = 24.22000 = 24.(210)200 ≡ 24 ≡ 5 (mod 11)
Vậy 20042004 chia 11 dư 5.
Ta có : 1944 ≡ -2 (mod 7) => 19442005 ≡ (-2)2005 (mod 7)
Mà (-2)3 ≡ - 1 (mod 7) => (-23)668 ≡ 1668 (mod 7) hay (-23)668 ≡ 1 (mod 7)
=> (-23)668.(-2) ≡ - 2 (mod 7) hay (-2)2005 ≡ - 2 (mod 7)
Vậy 19442005 cho 7 dư 5.
vì 2002 ⋮ 11 => 2004 - 2 ⋮ 11 => 2004 ≡ 2 (mod 11) => 2004^2004 ≡ 2^2004 (mod 11) , Mà 2^10 ≡ 1 (mod 11) vì 1024 - 1 ⋮ 11 => 2004^2004 = 2^4.2^2000 = 2^4.(2^10)^200 ≡ 2^4 ≡ 5 (mod 11) Vậy 2004^2004 chia 11 dư 5. 
