What is the last digit of $\operatorname{lcm}(3^{2003}-1,3^{2003}+1)$?
I am able to find out that LCM is $\dfrac{3^{4006}-1}2$. Since $3^{4006}$ has last digit as $8$, now second last digit can be anything from $0-9$. Based on that second last digit, my answer will vary. Please help how to go further?
On
$$3^{4006}=9^{2003}$$ $$=(10-1)^{2003}\equiv-1+\binom{2003}110+\binom{2003}210^2-\binom{2003}310^3+\cdots+10^{2003}$$ $$\equiv-1+2003\cdot10+\frac{2003\cdot2002}2\cdot10^2\pmod{200}$$
As $2003\equiv3,2002\equiv2\pmod {200}$ $$\implies 3^{4006}\equiv-1+3\cdot10+\frac{3\cdot2}2\cdot10^2\equiv329\equiv129\pmod{200}=200a+129$$ where $a$ is some integer
$$\implies \frac{3^{4006}-1}2=\frac{200a+129-1}2=100a+64\equiv64\pmod{100}$$
The two numbers have GCD equal to $2$. So, you're basically asking for the last digit of $(3^{2003}-1)(3^{2003}+1)/2 = (3^{4006}-1)/2$.
By Fermat's theorem, $3^{10 \cdot 4} \equiv 1 \pmod{100}$, since: $$\phi(100) = \phi(5^2 \cdot 2^2) = 5 \cdot 2 \cdot (5-1) \cdot (2-1) = 40 .$$
Thus, the two last digits of $(3^{4006}-1)$ are the same as those of $3^{6}-1 = 728$. Hence, the last digit of $(3^{4006}-1)/2$ is $4$, since you have no carry from the tens.