Prove by induction the following In the decimal form of the number $3^n$, the second from the end digit is even.
My proof so far:
Base Case: $n=3$
$3^3=27$. The second from the end digit is even, so the base case is true.
Inductive Step:
Assume $n=k$ is true and prove $n=k+1$ is true.
When $n=k+1$, $3^{k+1}=3(3^k)$
I don't know what to do now... the hint I was given was "What the end digit in a power of 3 can be? How does the second from the end digit
change when we multiply the number by 3?"
 A: I don't know if you like this answer. No need to use induction.
You'll see 
$$03,09,27,81,43,29,87, 61,83,49,47,41,23,69,07,21,63,89,67,01,03.$$
A: We will use the following lemma:

Lemma: For any $n \in \mathbb N$, the end digit of $3^n$ must be in the set $S = \{1,3,7,9\}$.

Proof: You can prove this by induction. There are four base cases, and for the induction hypothesis, you can consider each of the four cases separately. For example, if the last digit is a $7$, then multiplying by $3$ yields $21 \equiv 1 \pmod{10}$, which is in $S$.
As for the inductive step in your question, we consider the number $3^{k+1} = 3(3^k)$. Now by the above lemma, $3^k$ falls under $1$ of $4$ cases. Here's one of the cases.
Suppose that $3^k$ ends with a $9$. Now by the inductive hypothesis, the parity of the second from the end digit of $3^k$ is even, say $2a$. But when we multiply $3^k$ by $3$ (using elementary school multiplication), multiplying the ones digits yields $9 \cdot 3 = 27$, so the ones digit is now a $7$. Carrying the $2$, we see that the tens digit must be:
$$
3(2a) + 2 = 2(3a + 1)
$$
(after possibly having to take the remainder modulo $10$). So the second from the end digit of $3^{k+1}$ is still even, as desired.
A: $$3^4=81=20\cdot4+1\implies 3^{4n}=(20\cdot4+1)^n=1+\sum_{1\le r\le n}\binom nr20^r$$ which can be written as $20a+1$ where $a$ is a positive integer
In congruence term,   $\displaystyle 3^4\equiv1\pmod{20}\implies 3^{4n}\equiv1^n\equiv1$
Now, $\displaystyle 3^{4n+1}=3(20a+1)=60a+3$
$\displaystyle 3^{4n+2}=9(20a+1)=180a+9=20(9a)+9$
$\displaystyle 3^{4n+3}=27(20a+1)=540a+27=20(27a+1)+7$
Observe that
$\displaystyle(20b+m)(20c+n)=20(20bc+bn+cm)+mn$ will have the second from the end digit is even if $mn$ has the same property.
This is satisfied by $\displaystyle(m=1,n\in(1,3,5,7));(m=3,n\in(1,3,7)); (m=5,n\in(1,5));(m=7,n\in(1,3,7));$
So, we can safely conclude that the second from the end digit of $\displaystyle (20p+q)^r$ is even for $q\in(1,3,5,7)$ and $p,r$ are non-negative integers 
A: i do not know if this is a "proof" but, I like the question so thought of trying out : 
$3^3=27$
Induction step assumption is that $3^k$ has last but one digit from right is even.
$3^k=****(2a)K$
Now, last digits of powers of $3$ could be $1,3,7,9$.
Now, there would be no problem if $K$ in that $3^k$ is either $1$ or $3$ because multiplication by $3$ when we are considering $3^{k+1}=3.3^k$ does not affect the second term as
$3.1=3$ and $3.3=9$ so there will be no contribution to tens digit with this multiplication.
There will be a problem only if last digit is either $7$ or $9$
But then this can also be taken care as follows :
Suppose $K=7$ we would have :
$3.7=21\Rightarrow 3^{k+1}=3^k.3=(****(2a)7).3=****(2a+2)1=****(2b)1$ so last but one digit would be even.
Suppose $K=9$ we would have :
$3.9=271\Rightarrow 3^{k+1}=3^k.3=(****(2a)9).3=****(2a+2)7=****(2b)7$ so last but one digit would be even.
