Notation: $R(\frac{a}{b})$ means Remainder when a is divided by b. Now

$R(\frac{(2222)^{5555}}{7})$ $=R(\frac{3^{5555}}{7})$ $=R(\frac{(3^6)^{925}•3^5}{7})$ $=R(\frac{3^5}{7})=5$ And, $R(\frac{(5555)^{2222}}{7})$ $=R(\frac{4^{2222}}{7})$ $=R(\frac{(4^3)^{740}•3^2}{7})=2$

So that,

$R(\frac{(5555)^{2222}+(2222^){5555}}{7})=R(\frac{2+5}{7})=\boxed 0$

By Fermat's Little Theorem we have that both

$2222^{6} \equiv 1 \mod{7}$ and $5555^{6} \equiv 1 \mod{7}$ .

Now $5555 = 6*925 + 5$ and $2222 = 6*370 + 2$ ,

so we are now looking for $(2222^{5} + 5555^{2}) \mod{7}$ .

Next, note that $2222 = 7*317 + 3$ and $5555 = 7*793 + 4$ ,

and so we now have $(3^{5} + 4^{2}) \mod{7} \equiv 259 \mod{7} \equiv 0 \mod{7}$ ,

since $259 = 7*37$ . Thus the desired remainder is $\boxed{0}$ .

undefined value divided by any number yields zero

2222 mod 7 = 3 and 5555 mod 7 = 4 or -3 so (2222^5555)+(5555^2222) mod 7 = 3^5555 + (-3)^ 2222 mod 7 = 3^5555+ 3^2222 mod 7 = 3^2222(3^3333 + 1) mod 7 As 3^2222 is not divisible by 7 so we need to show that 3^3333 + 1 mod 7 = 0 Now as 7 is prime so as per format's little theorem 3^6 mod 7 = 1 3^3333 mod 7 = 3^(3333 mod 6) mod 7 = 3^ 3 mod 7 = 27 mod 7 So 3^3333 + 1 mod 7 = 28 mod 7 = 0

Final Answer is 0

In response of Sanjeet Raria's solution

As we know 2222 mod 7 = 3 and 5555 mod 7 = 4 or -3 so (2222^5555)+(5555^2222) mod 7 = 3^5555 + (-3)^ 2222 mod 7 = 3^5555+ 3^2222 mod 7 = 3^2222(3^3333 + 1) mod 7 As 3^2222 is not divisible by 7 so we need to show that 3^3333 + 1 mod 7 = 0 Now as 7 is prime so as per format's little theorem 3^6 mod 7 = 1 3^3333 mod 7 = 3^(3333 mod 6) mod 7 = 3^ 3 mod 7 = 27 mod 7 So 3^3333 + 1 mod 7 = 28 mod 7 = 0.