Is the ten's digit even in any power of 20n+c, where c is an odd digit? 
Prove or disprove that in any power of $20n+c$, where $c$ is an odd digit (i.e., $1,3,5,7,9$), the ten's digit is even.

This is probably a generalization of this.
I tried in the following way.
I observe that $(20m+c)(20n+d)$ is $20(20mn+m+n)+cd \equiv cd \pmod{20}.$
First of all, powers of $5$ end with $5$.
So, putting $c=d=5$, we get $25$, or directly
$$(20a+5)(20b+5)=20(20ab+a+b+1)+5,$$
clearly the ten's digit is even.
Now the product of any two numbers in $S=\{1,3,7,9\}$ is of the form $20r+e$ where $r$ is a positive integer and  $e \in S$.
If it obvious that the powers of any number ending with $e \in S$, will end with a member of $S$, we are done.
Any rectification or better solution is more than welcome.

I was trying to take the modulo 20 on the product of each member of $T=\{10a+b\}$ with each of the member of $U=\{10c+d\}$ where a,c are natural numbers, b,d are odd digits. I observed that the tens' digit is even when x,y are of same parity(for example, 11*11=121, 11*13=143, 11*17=187, 13*13=169, 13*17=221 etc.) except the cases of 3*5 and 5*7 where x,y need to be in the opposite parity to make the tens' digit even(for example, 13*15=195, but 13*25=325; 15*17=255, but 15*27=405 etc.). This is a further generalization from the power of a number to the product of different numbers and the numbers are of the form 10n+b instead of 20n+b.
 A: Suppose $(20n+c)^k = 20\alpha + c^k$ where $\alpha \in \mathbb{N} \cup \{ 0 \}$ and $c$ is an odd digit. (This certainly holds for $k=1$.)
Then
$$\begin{align}(20n+c)^{k+1} &= (20\alpha + c^k)(20n+c) \\ &= 20(20\alpha n + \alpha c + nc^k) + c^{k+1}\end{align}$$
So we've proved by induction that all powers of $20n+c$ are of the form $20\alpha + c^k$.
It suffices to show that the powers of $1,3,5,7,9$ less than $100$ have even tens digit. This is something you can verify directly. Indeed, they are $1, 3, 9, 27, 81, 5, 25, 7, 49, 9, 81$, and so we're fine.
A: $$(20n+c)^m =\sum_{k=0}^m \binom{m}{k} (20n)^{k}c^{m-k} \equiv \binom{m}{k} 20nc^{m-1} + c^m  \mod 20 $$
So the problem reduces to proving that the 10's digit of $c^m$ is even. This follows from a simple lemma.
Lemma Let $c$ be an odd digit, and let $k>1$ be so that $c^k \equiv c \mod 20$. If the tens digits of $c, c^2,..., c^{k}$ are even, then the tens digit of $c^m$ is even for all $m$.
This lemma is easy to prove (Prove it) and to complete the exercise you can observe that
$$1^2\equiv 1 , 3^5 \equiv 3, 5^2 \equiv 5, 7^5 \equiv 7, 9^3 \equiv 9 \mod 20 \,.$$
