I just want to know a good way to attack problems of this sort. I think I have the idea somewhat, however some expertise on the matter could go a long way in hammering home these ideas.
What is the last digit of 2^1000?
I have it as being 6 because 2^1000 is also 2^(2^3)(5^3)which is simply 2^8(125) which is 256^125. Any 6 when put to any power will always have a six as the last digit. That is how I got my answer but I am sure that there is a much more formal and proper way to write this out. Thanks a lot in advance for any help on this.