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.

A slightly different approach:
Let $n$ be the exponent of the variable power of $3$. Therefore, the power of $3$ takes the form $3^n$. Now, $n$ can either be even or odd. Hence two cases are formed:-
Case1: $n$ is even.
 Since $n$ is even, let us say it is of the form $2k$.
Therefore, $3^n = 3^{2k} = (3^k)^2$. This means that is $n$ if even, $3^n$ is a square. Notice that the even powers of $3$ result in a number with the last digit $\in \{1,9\}$
And $(2m + 1)^2 \equiv 1(mod 4)$
$\Rightarrow 3^n \equiv 1(mod4)$
$\Rightarrow 3^n -1 \equiv 0(mod4)$
$\Rightarrow 4 |(3^n-1)$.
We know that, if $4$ divides a number, it must divide $(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)$
Therefore, if $4 |(3^n-1)$,$ $ $ $ $ $ $ $ $ $ $ $   $4|(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)-1$
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 1-1)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 9-1)$.
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 0)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 8)$.
Since, $4| 0$ and $4| 8$, $4|(10($$the$ $second$ $last$ $digit$$)$
This is true only if $the$ $second$ $last$ $digit$ is even.
Case2: $n$ is odd.
 Since $n$ is odd, let us say it is of the form $2k+1$.
Therefore, $3^n = 3^{2k+1} = (3^k)^2.(3)$. This means that if $n$ is odd, $3^n$ is a $($square $* 3)$. Notice that the odd powers of $3$ result in a number with the last digit $\in \{3,7\}$
And $(2m + 1)^2 \equiv 1(mod 4)$
$\Rightarrow (2m + 1)^{2p+1} \equiv (2m + 1)(mod4)$
$\Rightarrow 3^n \equiv 3(mod4)$
$\Rightarrow 3^n -3 \equiv 0(mod4)$
$\Rightarrow 4 |(3^n-3)$.
We know that, if $4$ divides a number, it must divide $(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)$
Therefore, if $4 |(3^n-3)$,$ $ $ $ $ $ $ $ $ $ $ $   $4|(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)-3$
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 3-3)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 7-3)$.
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 0)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 4)$.
Since, $4| 0$ and $4| 4$, $4|(10($$the$ $second$ $last$ $digit$$)$
This is true only if $the$ $second$ $last$ $digit$ is even.
Therefore, the tens digit of any power of $3$ is always even.
