What is $(7^{2005}-1)/6 \pmod {1000}$? What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$
My approach:
Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.
So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.
Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.
Any help is appreciated!

We must multiply the modulus by $\,6\,$ to balance the division by $6$, i.e.
$\qquad 6\mid a \,\Rightarrow\,  a/6 \bmod 1000 = (a \bmod 6000)/6\ $ by the mod Distributive Law
$6000\!=\! 2^4\cdot 3\cdot 5^3$  whose totients $2^3,2,100\mid 2000\,$ so $\,\color{#c00}{7^{2000}\!\equiv 1}\bmod 2^4,3,5^3$ so also mod $6000,\,$ so $\bmod 6000\!:\ a\equiv 7^5 \color{#c00}{7^{2000}}-1 \equiv 7^5-1\equiv4806,\,$ so $\bmod 1000\!:\ a/6\equiv 4806/6 \equiv 801$.
