Fuhsaha, I think that's correct, but the question asked for the tens digit.
1) Since 19 is prime, Fermat's Little Theorem is applicable:
\(5^{400} = 5^{22(18)+4}\\ 5^{22(18)+4} {\displaystyle \equiv } 5^4 (\text{mod 19})\\5^4 = 625\\625{\displaystyle \equiv }17(\text{mod 19})\)
This means the remainder is \(\boxed{17}\)
2) I know there is a way of doing this using pure modular arithmetic by taking that number mod 100, but I still don't really know how to do it. Here's the look-for-cycle method just like fuhsaha used:
7^1 = 07 (last 2 digits)
7^2 = 49
\(\)7^3 = 43
7^4 = 01
7^5 = 07
(by the way, you don't actually need to compute 7^3, 7^4, or 7^5, you just need to multiply the last 2 digits of the previous power by 7.)
That means there is a cycle of 4, and the last 2 digits of 7^2018 is 49, which means that the 10's digit is \(\boxed{4}\)
3) where is k?
Looks like a lot of idle-time work!
