Proof using strong induction a conjecture about $4^n$ Compute $4^1$, $4^2$, $4^3$, $4^4$, $4^5$, $4^6$, $4^7$, and $4^8$. Make a conjecture about the units digit of $4^n$ where $n$ is a positive integer. Use strong mathematical induction to prove your conjecture. 
Solution:
$4^1$ = $4$
$4^2$ = $16$
$4^3$ = $64$
$4^4$ = $256$
$4^5$ = $1024$
$4^6$ = $4096$
$4^7$ = $16384$
$4^8$ = $65536$
Conjecture: For $n>0$, the unit digit of $4^n$ is $4$, if $n$ is odd and is $6$, if $n$ is even.


*

*Basis:
For $n = 1 \to 4^1 = 4$
For $n = 2 \to 4^2 = 16$
The conjecture is right for n = 1 and n = 2

*Inductive step: 
$4^{k+1} = 4\cdot 4^k$
If $k$ is odd ($k+1$ is even), the unit digit of $4^k$ is $4$, but if we multiply a number that the unit digit is $4$ with a number that the unit digit is $6$ we get a number that the unit digit is $6$, because $4\times 4 = 16$.
If $k$ is even ($k+1$ is odd), the unit digit of $4^k$ is $6$, but if we multiply a number that the unit digit is $6$ with a number that the unit digit is $4$ we get a number that the unit digit is $4$, because $ 6\times 4 = 24$.
So, the unit digit of $4^{k+1}$ is $4$, if $n$ is odd and is $6$, if $n$ is even.

To be proven: If $n$ is a positive integer, the units digit of $4^n$ is $4$ if $n$ is odd and $6$ if $n$ is even.
Proof by strong induction: (Base cases $n=1$ and $n=2$.)
Now assume that $k\ge 3$ and the result is true for all smaller positive values of $k$. The goal is to prove the theorem for $n=k$.
Let $\ell =k-2$; note that $\ell\ge 1$.
If $k$ is odd, then $\ell$ is odd, and strong induction implies that the units digit of $4^\ell$ is $4$. Then the units digit of $4^k$ is the units digit of $4^\ell \cdot 4^2$, which is the units digit of $4 \cdot 16$, which is $4$.
If $k$ is even, then $\ell$ is even, and strong induction implies that the units digit of $4^\ell$ is $6$. Then the units digit of $4^k$ is the units digit of $4^\ell \cdot 4^2$, which is the units digit of $6 \cdot 16$, which is $6$.
Thus, we have proven that the units digit of $4^k$ is $4$ if $k$ is odd and $6$ if $k$ is even. This proves the result.
