A Remainder Question

What is the remainder when

2222^5555 + 5555^2222

Is divided by 8?

A Solution:

Firstly, 2222 = 277 x 8 + 6 and
5555 = 694 x 8 + 3.

So 2222^5555 + 5555^2222 has the “same” remainder when divided by 8 as 6^5555 + 3^2222.

Now consider the “remainder” when powers of 6 are divided by 8 & look for a pattern

6^1 = 6 (mod 8)
6^2 = 4 (mod 8)
6^3 = 0 (mod 8)
6^4 = 0 (mod 8)
And eventually, 6^5555 = 0 (mod 8) too.

Consider now 3^2222 when divided by 8:

3^1 = 3 (mod 8)
3^2 = 1 (mod 8)
3^3 = 3 (mod 8)
3^4 = 1 (mod 8) And so on,

That is,

3^(odd power) = 3 (mod 8)
3^(even power) = 1 (mod 8)

Hence, 3^(2222) = 1 (mod 8)

Thus, finally

2222^5555 + 5555^2222
= 6^5555 + 3^2222 (mod 8)
= 0 + 3^2222 (mod 8)
= 3^2222 (mod 8)
= 1 (mod 8)
= remainder of 1 when divided by 8,

Thus 1 (Answer)