If $x$ be the $L.C.M.$ of $3^{2002}-1$ and $3^{2002}+1$, then find the last digit of x Let $x$ be the $L.C.M.$ of $3^{2002}-1$ and $3^{2002}+1$, then find the last digit of x.
Could someone give me slight hint hint with this question?
 A: HINT
The only common divisor is $2$.
Check modulo 10. Use the fact that $3^4 = 1 \mod 10$
A: The least common multiple must be a multiple of $3^{2002} + 1$. Since $3^{2002} + 1$ is divisible by $10$, so is the L.C.M., and hence the last digit of the L.C.M. is $0$.
$3^{2002} + 1$ is divisible by $10$ since we have that
$$ 3^{2002} + 1 \equiv 9^{1001} + 1 \equiv (-1)^{1001} + 1 \equiv -1 + 1 \equiv 0 \pmod{10}.$$
A: Lat $a=3^{2002}-1$, $b=3^{2002}+1$. $\gcd(a,b)=2\;$ and
$$\gcd(a,b)\operatorname{lcm}(a,b)=ab=9^{2002}-1\equiv (-1)^{2002}-1\equiv 0\mod 10,$$
hence $\;\operatorname{lcm}(a,b)\equiv 5\mod 10$.
A: Note that gcd divides their difference $= 2,\,$ hence the gcd $ = 2$ since both are even. Therefore their lcm $L$ is their product $P$ over $2$. To compute $L = P/2\,$ mod $10$ we compute $P$ mod $20$
$${\rm mod}\ 20\!:\,\ P = 9^{2002}\!-1 = \color{#c00}{81}^{1001}\!-1 \equiv {\color{#c00}{\bf 1}}^{1001}\!-1\equiv 0\ \ {\rm by}\ \ \color{#c00}{81\equiv \bf 1}$$
Therefore $\ 20\mid P\,\Rightarrow\, 10\mid P/2 = L,\,$ so the lcm $L$ has last digit $= 0$
