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.

We know that
$3^1=03$
$3^2=09$
$3^3=27$
$3^4=81$
And so on.
Here we noticed that 81 is the largest two digit number that is in the form of $3^n$.
After that 3 digit number starts .
But till 81 if you see tens digit is even so here 2 case arises
Case 1
When we have last digit of $3^n$ = 3, 1
After multiplying it by 3 we see it not makes any effect on the ten's digit
and as ten's digit is even then any thing multiplied to even will give you even
So for this case ten's digit is even.
Case 2
When the last digit of $3^n$ = 9 , 7
Here after multiplying by 3 it will have a effect on ten's digit
But $9*3=27$
And$3*7=21$
Here wee see 2 will be added to ten's digit
Now again as before I said ten's digit is even then the product after multiplying it by 3 will be even and as 2 is added to the result it will also be an even number as 2 is even and even added to even gives us even.
Here's what I thought.
