Finding the unit digit How can I find the unit digit of $2 ^{9^{100}}$. Is there any general method of finding the unit digit?
 A: I'm not fully sure what the term "unit digit" translates to, as there is no wikipedia page for it. However, I will assume that this translates into the last digit of a number. 
Let $f(n)$ denote the unit digit of $n$. 
Note that $f(2^1)=2, f(2^2)=4, f(2^3)=8, f(2^4)=6$. Then note $f(2^5)=2, f(2^6)=4, f(2^7)=8, f(2^8)=6$. 
As you can see, the unit digits repeat themselves with a period of $4$. 
Then note $9 \equiv 1  \pmod 4$. We can conclude $9^{100}\equiv 1^{100} \pmod 4$. 
Thus, since the remainder of $9^{100}$ divided by $4$ is $1$, we conclude the last digit is $2$. 
A: If you look at $2^1, 2^2, 2^3 ...$ you will quickly notice the unit digits follow the pattern $2,4,8,6,2,4,8,6...$ to prove that this pattern will continue to repeat note that when we multiply a number by $2$ it is the unit digit alone that determines what the unit digit of the answer will be. Using this pattern we get that $2^9$ ends in $2$ (because the remainder when we divide $9$ by $4$ is $1$).  But this means that the unit digits of ${2^9}^i$ follow the same pattern and powers of $2$ and hence ${2^9}^9$ will also end in $2$! By the same reasoning ${{2^9}^9}^9$ will also end in $2$ and so forth. And hence ${2^9}^{100}$ also ends in $2$ 
A: Note that the positive integer powers of $2$ go $2,4,8,16,32,...$. The unit digit goes in a cycle of length $4$.
So you need to determine $9^{100}\pmod 4$
Note that $ 9 = 8+1$. A simple application of binomial theorem will lead you to the answer.
A: The unit digit $\implies\pmod{10}$
Now as $(2,10)=2$ let us find $2^{9^n-1}\pmod{10/2}$
As $\phi(5)=4$ and $9\equiv1\pmod4$
$9^n\equiv1^n\equiv1\implies9^n-1\equiv0\pmod4$
$\implies2^{9^n-1}\equiv2^0\pmod5$
$\implies2^{9^n-1}\cdot2\equiv2^0\cdot2\pmod{5\cdot2}$
A: Yes, there's a general method: computing the number modulo $10$.
Observe the successive positive powers of $2\mod 10$ is periodic:
$$ 2^n\equiv\begin{cases}
2 &\text{if }\;n\equiv 1\mod 4,\\ 4  &\text{if }\;n\equiv 2\mod 4, \\8  &\text{if }\;n\equiv 3\mod 4, \\ 6  &\text{if }\;n\equiv 0\mod 4, 
\end{cases} $$
Thus $\;2^{9^{100}}\equiv 2^{9^{100}\bmod 4}\mod 10 =2^{(9\bmod 4)^{100}}=2^{1}=2\mod 10 .$
A: Hint $ $ Examining $2^n$ mod $10$ gives evidence for the conjecture that it has period $\,4\,$ for $\,n\ge 1.\,$ ${\rm mod}\ 4,\,$ the power we seek is $\,9^{100}\equiv 1^{100} \equiv \color{#c00}{\bf 1}.\,$ Low powers hint $\,\color{#c00}{2^{4n+\bf 1}\!\equiv 2}\pmod{10}\,$ so we rigorously prove this conjecture by induction. Base case is  $\,10\mid \color{#0a0}{2^{\large 5}-2},\,$ and inductive step 
$\quad {\rm mod}\ 10\!:\,\ \color{#c00}{2^{\large 4n+1}\equiv 2}\,\Rightarrow\,2^{\large 4(n+1)+1}\!\equiv 2^{\large 4} \color{#c00}{2^{\large 4n+1}}\!\equiv 2^{\large 4}\color{#c00}2\equiv \color{#0a0}{2^{\large 5}\equiv 2}.\ \ $  
Remark $\ $ If modular arithmetic is unfamilar you can eliminate it above to obtain
$$ 2^{\large 4(n+1)+1}\!-2 = 2^4(\color{#c00}{2^{4n+1}\!-2}) + \color{#0a0}{2^5 - 2} $$
By our induction hypothesis we know that  $\,10\mid \color{#c00}{2^{4n+1}-2}\,$ and $\,10\mid \color{#0a0}{2^{\large 5}-2}\,$ by the base case, hence $10$ divides the RHS above, so also the equal LHS, which completes the inductive proof. 
Alternatively you can prove by induction that $\,2^4-1\mid 2^{4n}-1\,$ then scale that by $\,2,\,$ or you could instead make the substituttion $\,a=2^4\,$ in $\,a-1\mid a^n-1\ $ (Factor Theorem).
More generally a very similar proof shows that $\,30\mid a^{4n+1}-a\ $ for all $\,a.\ $
