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 $4$?

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 $2$).
