Prove that $7^{100} - 3^{100}$ is divisible by $1000$ 
Prove that $7^{100} - 3^{100}$ is divisible by $1000$

Equivalently, we want to show that $$7^{100} = 3^{100} \pmod {1000}$$
I used WolframAlpha (not sure if that's the right way though) and found that $\varphi (250) = 100$.
So by Euler's theorem: $$7^{100} \equiv 7^{\varphi(250)} \equiv 1 \pmod {250} \\ 3^{100} \equiv 3^{\varphi(250)} \equiv 1 \pmod {250}$$
but of course, we want $\pmod {1000}$.
Is that what I'm intended to do in this exercise (how to proceed if so)? Is there a solution without the need to use WolframAlpha?
Thanks!

\begin{eqnarray} 
7^{100}-3^{100} &=& (10-3)^{100}-3^{100}\\
 &=& \underbrace{{100\choose 0}10^{100}-{100\choose 1}10^{99}\cdot 3+...-{100\choose 97}10^3 \cdot 3^{97}}_{10^3k}+{100\choose 98}10^2 \cdot 3^{98} -{100\choose 99}10 \cdot 3^{99}+3^{100}-3^{100}\\
&=&1000k +50\cdot 99\cdot10^2 \cdot 3^{98} -100\cdot 10 \cdot 3^{99}
\end{eqnarray} 
