We observe that:

2222(Mod 7) = 3, and:
5555(Mod 7) = 4

Thus,

(2222^5555 + 5555^2222)(Mod 7)
=(3^5555 + 4^2222)(Mod 7)
= ((3^5)^1111 + (4^2)^1111)(Mod 7)
= (243^1111 + 16^1111) (Mod 7)
= ((-2)^1111 + 2^1111) (Mod 7)
= (-2^1111 + 2^1111) (Mod 7)
= 0

Consequently, (2222^5555 + 5555^2222) is divisible by 7

Edited on November 14, 2007, 4:13 am

Posted by K Sengupta on 2007-11-14 04:12:54