gcd($3^{2003}+1\ , 3^{2003}-1$) $=gcd(3^{2003}+1\ , 2)\ as\ gcd(a,b)=gcd(a,a-b)\ and\ 3^{2003}+1\ -(3^{2003}-1)=2$ =2 as $3^n-1$ is even for any natural number n.
So,lcm($3^{2003}+1\ , 3^{2003}-1$)= ($3^{2003}+1)(3^{2003}-1$)/gcd($3^{2003}+1,\ 3^{2003}-1$)=($3^{4006}-1)/2$
Now, using Euler's Totient Theorem, $a^\phi(m)≡1(mod\ m)$ where (a,m)=1
So, $a^{k\phi(m)+h}≡a^h(mod\ m)$
Here (3,10)=(3,100)=1
Now, $\phi(10)=phi(2)phi(5)=1*4=4\ and\ \phi(100)=10phi(10)=40$
4006≡6(mod 40)≡2(mod 4)
$3^{4006}≡3^6(mod\ 100)≡3^2(mod\ 10)$
So, $lcm(3^{2003}+1\ , 3^{2003}-1)≡(3^6-1)/2(mod\ 100)≡(3^2-1)/2(mod\ 10)≡14(mod\ 100)≡4(mod\ 10)$