Last digits

Since 6 and 100 are not coprime integers, let us consider $6 \text{ mod }4$ and $6 \text{ mod }25$ separately.

$\begin{aligned} 6^{2007} & \equiv 6^{2 \times 1002}6^3 \equiv 36^{1002}6^3 \equiv 0 \text{ mod }4 \end{aligned}$

$\begin{aligned} 6^{2007} & \equiv 6^{\color{#3D99F6} 2007 \text{ mod }\phi (25)} \text{ (mod 25)} & \small \color{#3D99F6} \text{Since } \gcd (6,25) = 1 \text{, Euler's theorem applies.} \\ & \equiv 6^{\color{#3D99F6} 2007 \text{ mod }20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (25) = 20 \\ & \equiv 6^7 \text{ (mod 25)} \\ & \equiv 6^36^4 \text{ (mod 25)} \\ & \equiv 216(1296) \text{ (mod 25)} \\ & \equiv 16(-4) \text{ (mod 25)} \\ & \equiv -64 \equiv 11 \text{ (mod 25)} \end{aligned}$

Since $6^{2007} \equiv 0 \text{ (mod 4)}$ ,

$\begin{aligned} 4n & \equiv 11 \text{ (mod 25)} & \small \color{#3D99F6} \text{where }n \text{ is a positive integer.} \\ \implies n & = 34 \end{aligned}$

$\begin{aligned} 6^{2007} & \equiv 4n \equiv 4(34) \equiv 136 \equiv 36 \text{ (mod 100)} \end{aligned}$

Now we consider

$\begin{aligned} 7^{2007} & \equiv 7^{\color{#3D99F6} 2007 \text{ mod }\phi (100)} \text{ (mod 100)} & \small \color{#3D99F6} \text{Since } \gcd (7,100) = 1 \text{, Euler's theorem applies.} \\ & \equiv 7^{\color{#3D99F6} 2007 \text{ mod }40} \text{ (mod 100)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (100) = 40 \\ & \equiv 7^7 \text{ (mod 100)} \\ & \equiv 7^37^4 \text{ (mod 100)} \\ & \equiv 7^349^2 \text{ (mod 100)} \\ & \equiv 343(50-1)^2 \text{ (mod 100)} \\ & \equiv 43(1) \equiv 43 \text{ (mod 100)} \end{aligned}$

Therefore, $6^{2007} + 7^{2007} \equiv 36 + 43 \equiv \boxed{79} \text{ (mod 100)}$ .