Find the last Digit of $237^{1002}$? I looked at alot of examples online and alot of videos on how to find the last digit  But the thing with their videos/examples was that the base wasn't a huge number. What I mean by that is you can actually do the calculations in your head. But let's say we are dealing with a $3$ digit base Number... then how would I find the last digit. 
Q: $237^{1002}$
EDIT: UNIVERSITY LEVEL QUESTION. 
It would be more appreciated if you can help answer in different ways. 
Since the Last digit is 7 --> 


*

*$7^1 = 7$

*$7^2 = 49 = 9$

*$7^3 = 343 = 3$ 

*$7^4 = 2401 = 1$ 
$.......$
$........$

*$7^9 = 40353607 = 7$

*$7^{10} = 282475249 = 9$


Notice the Pattern of the last digit. $7,9,3,1,7,9,3,1...$The last digit repeats in pattern that is 4 digits long.


*

*Remainder is 1 --> 7

*Remainder is 2 --> 9

*Remainder is 3 --> 3

*Remainder is 0 --> 1


So, $237/4 = 59$ with the remainder of $1$ which refers to $7$. So the last digit has to be $7$.
 A: You want to know the last digit of $237^{1002}$, which is the same as the remainder of $237^{1002}$ after division by $10$. This calls for modular arithmetic. From $237\equiv7\pmod{10}$ it follows that
$$237^{1002}\equiv7^{1002}\pmod{10}.$$
Now the base number is small; can you take it from here?
A: $ {\rm mod}\ 10\!:\ \color{#c00}{7^{\large 4}\equiv\bf 1}\,\Rightarrow\, 7^{\large J+4K}\!\equiv 7^{\large J}(\color{#c00}{7^{\large 4}})^{\large K}\!\equiv 7^{\large J}\color{#c00}{\bf 1}^{\large K}\!\equiv 7^{\large J}\, $ by standard Congruence Rules.
Finally write $\ 1002 = J\!+\!4K\ $ for $\,0\le J < 4\ $  and apply the above.
A: We show that the last two digits of $237^{1002}$ are $69$.
For any $n \in \mathbb N$, the two digits of $n$ are given by $n\bmod{100}$. If $\gcd(n,100)=1$, $n^{\phi(100)}=n^{40} \equiv 1\pmod{100}$, by Euler's theorem.
Since $\gcd(237,100)=1$, applying Euler's theorem gives
$$ 237^{1002} = \big(237^{40}\big)^{25} \cdot 237^2 \equiv 237^2 \equiv 37^2 \equiv 69\pmod{100}. $$
This proves our claim. $\blacksquare$
A: Simple version without the notation:
$7 \times 1 = 7$
$7 \times 7 = 49$
$7 \times 9 = 63$
$7 \times 3 = 21$
Just look at the last digit in each case.
So the last digit of $7^1$ is $7$.  The last digit of $7^2$ is $9$.  The last digit of $7^3$ is $3$.  And, the last digit of $7^4$ is $1$.
Thus the last digit of $7^5$ is also $7$.  And the last digit of $7^9$ is $7$ (because $7^4 \times 7^4 \times 7 = 7^9$, and the last digits thereof are $1 \times 1 \times 7$.)
And the last digit of $7^{51}$ is the same as the last digit of $7^{47}$, which is the same as the last digit of $7^{43}$, which is the same as the last digit of $7^{39}$ (see the pattern?)...which is the same as the last digit of $7^{7}$, which is the same as the last digit of $7^3$, which is $3$.
By the same logic, the last digit of $7^{1002}$ is the same as the last digit of $7^2$, which is $9$.
A: There's a quite simple way of solving these kinds of problems using Chinese remainder theorem and Fermat's little theorem. 
We want to know $237^{1002}$ mod $10$. As Servaes has pointed out, $237^{1002} \equiv 7^{1002} \mod{10}$, so we can work with $7^{1002}$ which is simpler.
Using Fermat's little theorem (which tells us $7^4 \equiv 1$ (mod $5$)) we get:
 $7^{1002} \equiv (7^4)^{250}\cdot7^2 \equiv 1^{250}\cdot49 \equiv 49 \equiv 4\ \ (\text{mod } 5)$
Also $7^{1002} \equiv 1^{1002} \equiv 1\ \ (\text{mod } 2)$, 
The chinese remainder theorem tells us that the system
\begin{cases}
x \equiv 4 \ \ \text{(mod } 5) \\
x \equiv 1 \ \ \text{(mod } 2) \\
\end{cases}
Has exactly one solution mod $10$. It's easy to find it by trial and error since we only have 10 options: the solution is $x \equiv 9$  (mod $10$).
Treating $7^{1002}$ as our unknown and applying that result we conclude $7^{1002} \equiv 9$ (mod $10$).
Edit: a quicker but less mechanic way to solve this particular problem is to use Euler's theorem, which is a generalized version of Fermat's little theorem. 
Euler's theorem tells us that if $a$ and $n$ are relatively prime then $a^{\phi(n)} \equiv 1$ (mod $n$), where $\phi(n)$ counts the positive integers up to a given integer $n$ that are relatively prime to $n$. 
With $n = 10$ and $a = 7$ we have $7^4 \equiv 1$ (mod 10). (Of course we don't need the theorem to know this, given that $7^4 = 2401$, but that's where we get the idea). 
So $7^{1002} \equiv (7^4)^{250} \cdot 7^2 \equiv 1^{250}\cdot49\equiv49\equiv 9 $ (mod 10).  
A: Short version:
Modulo $10$, the powers of $237$ up to $1002$, like those of $7$, are $1,7,9,3,1,\cdots 9$. (The period is $4$).
A: It will be many words to show the related theroy.  So I would like the solve the problem as follow:


*

*$237 \equiv 7 \pmod{10}$

*$7 ^ 4 \equiv 1 \pmod{10}$

*$1002 \equiv 2 \pmod{4}$


Now $237^{1002} \equiv 7^{1002} \equiv 7 ^ 2 \equiv 49 \equiv 9 \pmod{10}$.
So your answer is $9$. In a similar way, you can get the last two digits is $69$ as follow.


*

*$237 \equiv 37 \pmod{100}$

*$37 ^ {20} \equiv 1 \pmod{100}$

*$1002 \equiv 2 \pmod{20}$


Now $237^{1002} \equiv 37^{1002} \equiv 37 ^ 2 \equiv 1369\equiv 69 \pmod{100}$.
A: $237^{1002}=237^{1000+2}$
$237^2 \times 237^{1000}$
as you know unit digit in case of powers repeat after each fourth power 
so $unit digit(237^2\times 237^{1000})=unit digit(237^2)\times unit digit(237^{1000})$
$=9\times 1=9$
A: $237\equiv-3\mod10$   
$\therefore 237^{1002} \equiv(-3)^{1002} \equiv (-3)^2.(9)^{500}\equiv 9.(-1)^{500}\equiv 9 \mod10$
A: Finding the last digit of ${237}^{1002}$ is same as finding the last digit of $ 7^{1002}$.
We see that
$7^1 = 7$;
$7^2 = 9$;
$7^3 = 3$;
$7^4 = 1$;
$\dots$
Then comes the repetition of $7$, $9$, $3$ and $1$ as the powers of $7$ increase.
Since our power is even, our choices reduce to last digit of $7^2$ and $7^4$ which are $9$ and $1$ respectively.
Since $1002$ is a multiple of $2$ but not a multiple of $4$,
our choice of power $1002$ of $7$ is same as power $2$ of $7$, which is $9$.
Therefore, $7^{1002} = 9$ (last digit)
A: You can solve this using Euler's theorem:


*

*$\gcd(237,10)=1$

*Therefore $237^{\phi(10)}\equiv1\pmod{10}$

*Therefore $237^{\phi(2\cdot5)}\equiv1\pmod{10}$

*Therefore $237^{(2-1)\cdot(5-1)}\equiv1\pmod{10}$

*Therefore $237^{1\cdot4}\equiv1\pmod{10}$

*Therefore $237^{4}\equiv1\pmod{10}$


Therefore
 $237^{1002}\equiv237^{4\cdot250+2}\equiv(\color\red{237^{4}})^{250}\cdot237^{2}\equiv\color\red{1}^{250}\cdot237^{2}\equiv56169\equiv9\pmod{10}$.
