In every power of 3 the tens digit is an even number How to prove that in every power of $3$, with natural exponent, the tens digit
is an even number?
For example, $3 ^ 5 = 243$ and $4$ is even.

It's actually interesting.
If you do a table of multiples of 1, 3, 7, 9 modulo 20, you will find a closed set, ie you can't derive an 11, 13, 17 or 19 from these numbers.  
What this means is that any number comprised entirely of primes that have an even tens-digit will itself have an even tens-digt.  Such primes are 3, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89, 101, 103, 107, 109  to a hundred.  
If the tens-digit is odd, then it must be divisible by an odd number of primes of the form 11, 13, 17, 19 mod 20.  Any even number of these would produce an even tens digit.
