How to compute that the unit digit of $\frac{(7^{6002} − 1)}{2}$? The mother problem is:

Find the unit digit in LCM of $7^{3001} − 1$ and $7^{3001} + 1$

This problem comes with four options to choose the correct answer from,my approach,as the two number are two consecutive even numbers hence the required LCM is  $$\frac{(7^{3001} − 1)(7^{3001} + 1)}{2}$$
Using algebra that expression becomes $\frac{(7^{6002} − 1)}{2}$,now it is not hard to see that unit digit of $(7^{6002} − 1)$ is $8$.
So the possible unit digit is either $4$ or $9$,As there was no $9$ as option I selected $4$ as the unit digit which is correct but as this last part is a kind of fluke I am not sure if my approach is right or not or may be I am unable to figure out the last part how to be sure that the unit digit of $\frac{(7^{6002} − 1)}{2}$ is $4$?
 A: Your argument is great! To finish it off, note that $\frac{7^{6002}-1}{2}=\text{LCM}(7^{3001}-1,7^{3001}+1)$ is even, so it could not have a units digit of 9.
A: Here is another way: $\ $ by Euler $\phi\:,\ $ $\rm\:7^{3001}\: \equiv\ 7\pmod{20}$   
thus $\rm\ n\: =\: (7^{3001}-1)\ (7^{3001}+1)\ \equiv\ 6\cdot 8\ \equiv\ 8\pmod{20}$  
Therefore, since $\rm\:n\:$ is even, we obtain $\rm\:n/2\ \equiv\ 4\pmod{10}\ $ by cancelling $2\:.$
NOTE $\ $ If Euler $\phi$ is unknown then one may proceed simply as follows
$\mod\ 20:\ \ 7^2 \equiv 9\ \Rightarrow\ 7^4 \equiv\ 9^2 \equiv 1\ \Rightarrow\ \ 7^{3001}\ \equiv\ 7\ (7^4)^{750}\: \equiv\ 7$
A: We look directly at the mother problem. Exactly as in your approach, we observe that we need to evaluate
$$\frac{(7^{3001}-1)(7^{3001}+1)}{2}$$
modulo $10$.
Let our expression above be $x$.  Then $2x= (7^{3001}-1)(7^{3001}+1)$.  We will evaluate $2x$ modulo $20$.
Note that $7^{3000}$ is congruent to $1$ modulo $4$ and modulo $5$. Thus  $7^{3001} \equiv 7\pmod{20}$, and therefore 
$$2x\equiv (6)(8)\equiv 8 \pmod{20}.$$
It follows that $x\equiv 4\pmod{10}$.
A: The elementary way: $7^2=50-1$ mod $100$ hence $7^4=1$ mod $100$ hence $7^{6000}=1$ mod $100$ because $6000$ is a multiple of $4$, hence $7^{6002}=7^2$ mod $100$ and your number is $\frac12(7^2-1)$ mod $50$. This is $24$ mod $50$ hence the last digit is $4$ (and a priori, the previous digit is either $2$ or $7$ but one just has to be a little more careful to prove that it is $2$).
