Find the last digit of $3^{1006}$ The way I usually do is to observe the last digit of $3^1$, $3^2$,... and find the loop. Then we divide $1006$ by the loop and see what's the remainder. Is it the best way to solve this question? What if the base number is large? Like $33^{1006}$? Though we can break $33$ into $3 \times 11$, the exponent of $11$ is still hard to calculate.
 A: You have $$3^2=9\equiv -1\pmod{10}.$$
And $1006=503\times 2$, so
$$3^{1006}=(3^2)^{503}\equiv (-1)^{503}\equiv -1\equiv 9\pmod{10}.$$
So the last digit is $9$.

And for something like $11$, you can use the fact that $11\equiv 1\pmod {10}.$
A: $3^{1006}$ or $33^{1006}$
doesn't really matter
$33\equiv 3\pmod {10}\\
33^{1006}\equiv 3^{1006}\pmod {10}$
$3^4 = 81$
You might say this as $3^4\equiv 1 \pmod{10}$
The last digit of $3^n$ is the same last digit as $3^{n+4k}$
that is:
$(3^{n+4k}) = (3^n)(3^{4k})\equiv (3^n)(1) \pmod{10}$
$1006 = 251\cdot 4 + 2$
the last digit of $3^{1006}$ is the same as the last digit of $3^2$
A: You have 
\begin{cases}3^1& =3\\ 3^2& =9 \\ 3^3&=27\\3^4&= 81\\3^5&= 243\end{cases} Thus the last digit repeats after $4$ steps. Since $1006=251\cdot 4+2$ it is
$$3^{1006}=(3^4)^{251}\cdot 3^2.$$ The last digit of $(3^4)^{251}$ is $1$ and the last digit of $3^2=9.$ So the answer you are looking for is $9.$ 
To compute the last digit of $33^{1006}$ you are in the right way. Since $33=3\cdot 11$ and the last digit of $11^n$ is $1$ you have that the last digit of $33^{1006}$ is the same of $3^{1006}.$
A: HINT:
Find the last digit of $3^{1006\bmod4}$.

Adding a formal solution, just in case anyone will find it useful:
You are looking for $3^{1006}\pmod{10}$.
Since $\gcd(3,10)=1$, by Euler's theorem, $3^{\phi(10)}\equiv1\pmod{10}$.
We know that $\phi(10)=\phi(2\cdot5)=(2-1)\cdot(5-1)=4$.
Therefore $3^{4}\equiv1\pmod{10}$.
Therefore $3^{1006}\equiv3^{4\cdot251+2}\equiv(\color\red{3^4})^{251}\cdot3^2\equiv\color\red{1}^{251}\cdot3^2\equiv1\cdot9\equiv9\pmod{10}$.
A: The powers of $3$ cycle from $1\to3\to9\to7$, depending upon the exponent's modulus with respect to $4$. Since $1006\equiv 2\pmod{4}$, the last digit  of $3^{1006}$ is $9$.
You can still use this tactic for larger bases. Suppose we want the last digit of $33^{1006}$, as you suggest. Since $33=30+3$, the powers of $33$'s last digit are completely determined by the powers of $3$.
A: Very soon you will learn Euler's theorem:
If the greatest common factor of $a$ and $n$ then $a^{\phi{n}} \equiv 1 \mod n$ where $\phi(n)$ is the number of numbers relatively prime to $n$ that are less than $n$.  
As $1,3,7$ and $9$ are relatively prime to $10$, and $\gcd(3,10) = 1$ we know $\phi(10) = 4$ and $3^4 \equiv 1 \mod 10$ so $3^{1006} = 3^{4*251 + 2} \equiv 3^2 \equiv 9 \mod 10$.
As $33 = 3*10 + 3$ $33^n = (30 + 3)^n = 10*something + 3^n$ will have the same last digit.  But $\gcd(33,10) = 1$ so $3^4 \equiv 1 \mod 10$.  And everything is the same.
What would be harder is the last two digits of $33^{1006}$.  $\gcd(33,100) =1$ so $33^{\phi(100)} \equiv 1 \mod 100$.  But what is $\phi(100)$?
There is a thereom that $\phi(p) = p-1$ and that $\phi(p^k) = p^{k-1}(p-1)$ and then $\phi(mn) = \phi(m)\phi(n)$ so $\phi(100)=\phi(2^2)\phi(5^2) = 2*1*5*4 = 40$. So there are $40$ numbers less than $100$ relatively prime to $100$.
and $33^{40} \equiv 1 \mod 100$ so $33^{1006 = 40*25 +6} \equiv 33^6 \mod 100 \equiv 30^2 + 2*3*30 + 3^2 \equiv 189 \equiv 89 \mod 100$.  The last two digits are $89$.
See:
https://en.wikipedia.org/wiki/Modular_arithmetic
https://en.wikipedia.org/wiki/Euler%27s_theorem
https://en.wikipedia.org/wiki/Euler%27s_totient_function
