A number theory problem by Madhavarapu Revanth

Since $10$ is coprime to $3$ , $7$ and $13$ , by Euler totient theorem we have $3^{\phi (10)} \equiv 7^{\phi (10)} \equiv 13^{\phi (10)} \equiv 1 \text{ (mod 10)}$ , where $\phi (10) = 4$ is Euler totient function. Therefore,

$\begin{aligned} 3^{2007}7^{2008}13^{2009} & \equiv 3^{4\times 501+3}7^{4 \times 502}13^{4 \times 502 + 1} \text{ (mod 10)} \\ & \equiv 27 \times 1 \times 13 \text{ (mod 10)} \\ & \equiv \boxed{1} \text{ (mod 10)} \end{aligned}$

$3^{2007} \times 7^{2008} \times 13^{2009} \equiv 3^{2007} \times (-3)^{2008} \times 3^{2009} \equiv 3^{2007 + 2008 + 2009} \equiv 3^{6024} \equiv 9^{3012} \equiv (-1)^{3012} \equiv 1 \mod 10$