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: Moron's answer is very elegant.
Here is an easy but tedious way of dealing with many similar questions: For any $n$ and $k$, the last $k$ digits of the numbers in the sequence $n^1,n^2,n^3,\dots$ eventually repeat, so we only need to check a few cases.
For the question at hand, note that $3^2=9=10-1$, so $3^{20}=(10-1)^{10}$ has the form $100k+1$ for some $k$. This means that the last 2 digits of powers of 3 repeat each 20 numbers. So we just need to check the last two digits of $3^0,3^1,\dots,3^{19}$, and this can be done very quickly even by hand: $$01, 03, 09, 27, 81, 43, 29, 87,\dots,89,67.$$
A: Use induction on the exponent:
The last digit is either 1, 3, 7 or 9.
If 1 or 3, multiplying a power of 3 will not change the evenness of the second last digit.
If 7 or 9, multiplying will result in a carry over of 2, which again does not change the evenness of the second last digit.
A: Hint $\,\bmod 20\!:\,\ 3^{\large 4} \equiv 1\,\Rightarrow\,3^{\large n}\equiv 3^{\,\large n\bmod 4}\in \{\color{#c00}01,\color{#c00}03,\color{#c00}09,\color{#c00}27\}$

Inductively: $\ 3^{\large n} = 20\,k\, +\{0,3,9,7\}\,\ $ 
$\qquad\ \Rightarrow\ 3^{\large n+1} = 20 K+\{0,9,7,1\},\,\  K = 3k\!+\!i,\,\ i\in \{ 0,1\}$ 
Notice $\,r< 10\,\Rightarrow\, 20k+r =  (\color{#0a0}{2k})10 + r\,$  has even tens digits $\,\color{#0a0}{2k} \bmod 10 = 2(k\bmod 5)$
A: 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.
A: 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.
A: 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.
A: Considering $3^k \bmod 20$, we know from the Carmichael function that the values will cycle on a pattern of $\lambda(20)=4$ (or less). This is short enough to just evaluate:
$
3^1\equiv 3 \bmod 20 \\
3^2\equiv 9 \bmod 20 \\
3^3\equiv 7 \bmod 20 \\
3^4\equiv 1 \bmod 20 
$
and the cycle repeats. As you can see, none of these values are greater than $10$ so the tens digit of any power of $3$ will be even (because these values will always be added to some multiple of $20$).
A: By brute force:
The powers of $3$ modulo $100$ are periodically
$$01,03,09,27,81,43,29,87,61,83,49,47,41,23,69,07,21,63,89,67,01,\cdots$$

By induction:
$$3^0\bmod20=1\in\{1,3,9,7\}$$
and
$$3^n\bmod20\in\{1,3,9,7\}\implies 3^{n+1}\bmod20\in\{3,9,7,1\}.$$
So the tenth digit modulo $2$ is always $0$.
