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.