Hello, padsinseven!
I think your solution is excellent.
It can be streamlined a bit . . .
What is the last digit of ?
Fact: any number ending in 6, when raised to a power, will end in 6.
Therefore, the last digit is 6.
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.
Another sobervation you could make is the last digit of the 2 powers.
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
2^7=128
2^8=256
2^9=512
2^10=1024
See, the last digit has the pattern: 2,4,8,6. Then it starts over.
Since 4 divides into 1000 evenly, the last digit is 6.
Suppose it were 2^2157112678. What's the last digit?.
4 divides 2157112678 with 2 as a remainder. Therefore, the last digit is 4.