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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 $4$?

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up vote 8 down vote accepted

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.

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Aha!!...missed just by a bit! Thank you very much :-) – Quixotic Aug 27 '11 at 17:05
Just another thing,could this problem be done any other faster way? – Quixotic Aug 27 '11 at 17:06
I can't think of any other approaches off the top of my head, but I'm sure there are plenty :) – Zev Chonoles Aug 27 '11 at 17:11
Or else, working directly with $7^{6002}$, use the fact that the square of an odd number is congruent to $1$ modulo $8$. – André Nicolas Aug 27 '11 at 17:12
@André Nicolas:That's a useful piece of fact,thanks :-) – Quixotic Aug 27 '11 at 17:14

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

$\mod\ 20:\ \ 7^2 \equiv 9\ \Rightarrow\ 7^4 \equiv\ 9^2 \equiv 1\ \Rightarrow\ \ 7^{3001}\ \equiv\ 7\ (7^4)^{750}\: \equiv\ 7$

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And, without enough theory to understand $\phi$, you can compute $7^x \mod 20$ for the first few values of $x$ to discover its periodicity. Thus easily get $7^{6002} \mod 20$, and then calculate your answer. – GEdgar Aug 27 '11 at 17:41
@GEdgar Of course one could say the same thing about every application of little Fermat or Euler $\phi$ for small modulus. At some point one needs to take the leap from ad-hoc specific instances to the general theorem. – Bill Dubuque Aug 27 '11 at 17:49
On the other hand one could say: Other linear recurrences, not just $a_{n+1} = 7 a_n$, can be handelled by the ad hoc method (discover periodicity) even if Euler/Fermat don't apply. – GEdgar Aug 27 '11 at 18:01
@GEdgar Sure, there are usually various ad-hoc approaches to elementary number theory problems, sometimes with advantages in other contexts. But for this class of problems said ad-hoc approach quickly becomes intractable once the modulus has nontrivial size. So I prefer to present the general approach when it appears that the OP may have knowledge of such. In any case I added a note to eliminate Euler $\phi$ if need be – Bill Dubuque Aug 27 '11 at 18:27

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 $x\equiv 4\pmod{10}$.

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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 $2$).

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