Finding the Units Digit to 7 to the 2945 How would I go about finding the Units Digit to 7^(2945)?
I know that:
7^0 = 1
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
...
7^9 = 40353607
 A: when u are interested only in the unit digit you can multiply step by step like u did in your post but you can forget the other digits.
So you have $$7^2 = 49$$ however you forget what was it exactly and u only remember that $$7^2 = x\cdot 10 + 9 $$
you multiply again
$$7^3=7^2 \cdot 7 = 70x+63 = (7x+6) \cdot 10+3$$
yet again you only remembering that it looked like
$$7^3=x_3 \cdot 10 +3$$ 
as you can see you dont have to remember $7^9 = 40353607$, $7^9 =x_9 \cdot 10 +7$ will be enougth
so now we can repeat you what u did but make it simpler. We can note that


*

*if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 7

*if the $7^n$ ends with $7$ then $7^{n+1}$ ends with 9

*if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 3

*if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 1


so the sequence of units digits will be $(1,7,9,3,1,7,3,9,\ldots)$
now you want 2945th element of that sequence. The value repeats itself evey 4 steps.$$a_n=a_{n-4}$$
and also
$$a_n=a_{n-20}$$
$$a_n=a_{n-100}$$
so u can say that:
$$a_{2945}=a_{45}=a_5=a_1=7^1=7$$
A: as you know $7^4$ is a number whose units digit is 1. You can write $7^{2945}$ as $7^{(736\times 4+1)}$ or alternately as $(7^4)^{736}7^1$ as the first factor when simplified  will be a number whose units digit is 1 so the answer is a number ending with "7" as its unit digit...
A: It is easy to verify that$$7^{4n+k}\equiv 7,9,3,1\pmod{10}\text{  when }k=1,2,3,4 \text{ respectively}$$ Hence
$$7^{2945}=7^{4\cdot736+1}=(7^{736})^4\cdot 7\equiv 7\pmod{10}$$
