Question on modulo Find the last two digits of $3^{2002}$.
How should I approach this question using modulo? I obtained 09 as my answer however the given answer was 43. 
My method was as follows:
$2002\:=\:8\cdot 250+2$
$3^{2002}=\:3^{8\cdot 250+2}\:=\:3^{8\cdot 250}\cdot 3^2$
The last two digits is the remainder when divided by 100. Thus we need to compute the sum mod 100.
$3^8\equiv 61\:\left(mod\:100\right)$
Thus, 
$3^{2002}=\:3^{8\cdot 250+2}\:=\:3^{8\cdot 250}\cdot 3^2$
$\equiv 61^{250}\cdot 9\:\equiv 61^{5\cdot 50}\cdot 9\equiv \:9\:\left(mod\:100\right)$
Therefore my answer would be 09. Am I doing it the right way? 
Thanks in advanced

Observe that $\text{gcd(3,100)} = 1$, by Euler totient formula: $3^{\phi(100)} \cong 1(\mod 100)$. But $\phi (100) = 100\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{5}\right) = 40$. Thus: $3^{40} \cong 1(\mod 100)$. So $3^{2002} = (3^{40})^{50}\cdot 3^2 \cong 1\cdot 9(\mod 100) \cong  9(\mod 100)$. Thus the remainder is $9$ as claimed.
