A number theory problem by Sam Sung

We have

$2222^{5555}+5555^{2222}$

$=\left(2222^5\right)^{1111}+\left(5555^2\right)^{1111}$

$=\left(2222^5+5555^2\right)\left[\left(2222^5\right)^{1110}-\cdots+\left(5555^2\right)^{1110}\right]$

Note that

$2222^5+5555^2$

$\equiv 3^5+4^2 \pmod{7}$

$\equiv 259 \pmod{7} \equiv 0 \pmod{7}$

Hence, $2222^{5555}+5555^{2222}$ leaves a remainder of $\boxed{0}$ when divided by $7$ , and we are done.

2222^5555 + 5555^2222 = (7a+3)^5555 + (7b+4)^2222 after removing all multiples of 7 from binomial expansion = 3^5555 + 4^2222 =(3^3)^1851 3^2 + (4^3)^740 4^2 =27^1851 * 9 + 64 ^ 740 * 16 = (7 4-1)^1851 * 9 + (7 9 +1 )^740 * 16 after removing all multiples of 7 from binomial expansions = (-1)^1851 * 9 + (1)^740 * 16 = -9 + 16 = 7... 7 still remains... thus term is always a multiple of 7........