What is the remainder when 5555^2222 + 2222^5555 is divided by 7?

What is the remainder when \(5555^{2222}+2222^{5555}\) is divided by 7?

In the binomial theorem, \((a + b)^n=(a^n) + nC1(a^{n-1}*b^1) + nC2(a^{n-2}*b^2) + .... + nC1(a^1*b^{n-1}) + b^n\).(C stands for the combination)
When it comes to the expression given, \(5555^{2222} = (7n + 4)^{2222}\). (n is the quotient of 5555/7)
Let's say a=7n and b=4, \((7n + 4)^{2222} = 7n^{2222} + ..... + (2222C2221)(7n^1 * 4^{2221}) + 4^{2222}.\)
Thus, \(5555^{2222} = 7m + 4^{2222}\).(7m is the simplified form of \(7n^{2222} + ..... + (2222C2221)(7n^1 * 4^{2221})\))
Similarly, \(2222^{5555} = (7r + 3)^{5555}\).
Hence, \(2222^{5555} = 7s + 3^{5555}\).
So, \(5555^{2222} + 2222^{5555} = 7m + 7s + 4^{2222} + 3^{5555}\)

Since we need only the remainder of the expression, we can ignore 7n.

Then, \(4^{2222} + 3^{5555} = 16^{1111} + 243^{1111}\)
Because \(a^n + b^n\) is divisible by a + b when n is odd.

\(4^{2222} + 3^{5555} = t * (16 + 243)\)
Because 259 is divisible by 7, the remainder is 0.

The answer is A

Mansoor50
Sorry, but I could not post a new comment.
So, I revised my post to express the binomial theorem explicitly.
Hope it helps.