2222^5555 + 5555^2222 = 3^5555 + 4^2222 (mod 7).
There are only 7 residue classes modulo 7. Therefore, the series of powers of a given number (modulo 7) is bound to be cyclic (of period 6).
For example, 3^2 = 2 (mod 7), 3^3 = 6 (mod 7), 3^4 = 4 (mod 7), 3^5 = 5 (mod 7), 3^6 = 1 (mod 7), 3^7 = 3 (mod 7), and so on: 3,2,6,4,5,1,3,...
Note that 5555 = 5 (mod 6). Therefore, 2222^5555 = 3^5 (mod 7) = 5(mod 7).
Similarly, 5555^2222 = 4^2222 (mod 7). Powers of 4 modulo 7 form a cycle of period 3: 4,2,1,4,... Hence, 4^2222 (mod 7) = 4^{2222( mod 3)}(mod 7) = 4^2 (mod 7) = 2 (mod 7).
Adding up 5 + 2 gives a number divisible by 7. So is 2222^5555 + 5555^2222.
Divisible by 7
