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**