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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.

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    $\begingroup$ It's a good way to solve it. The loop is pretty short. But if you know Eulers formula or fermats little formula you don't have to "find" the loop. You can calculate it with certainty. At any rate $3^2 \equiv -1 \mod 10$ so $3^4 \equiv 1 \mod 10$ so the loop is 4 and $1006$ has remainder $2$ so... it's not hard. $\endgroup$ – fleablood Dec 7 '16 at 21:09
  • $\begingroup$ Are you familiar with modular arithmetic, i.e. congruences, e.g. $\,3^2\equiv -1\pmod{10}\,?\ \ $ $\endgroup$ – Bill Dubuque Dec 7 '16 at 21:19
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    $\begingroup$ I think this falls under this umbrella question explaining how to do this and related problems. Please search the site for similar questions before asking. $\endgroup$ – Jyrki Lahtonen Dec 7 '16 at 21:39
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    $\begingroup$ See also these posts: Find the last two digits of $7^{81}$ and Find the last two digits of $3^{45}$. And perhaps other posts linked there $\endgroup$ – Martin Sleziak Dec 8 '16 at 3:09
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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 $11\equiv 1\pmod {10}.$

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$3^{1006}$ 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 $3^2$

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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 $3^{1006}.$

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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 $3^{1006}\equiv3^{4\cdot251+2}\equiv(\color\red{3^4})^{251}\cdot3^2\equiv\color\red{1}^{251}\cdot3^2\equiv1\cdot9\equiv9\pmod{10}$.

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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 $3$.

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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

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