Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem

What is the remainder when $10^{35}$ is divided by $7$? 

I used the binomial formula: $\dfrac {(7+3)^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot 3^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot + 35 \cdot 3^{34} \cdot 7}{7}  + \dfrac {3^{35}}{7}$
Therefore, $10^{35}$ will have the same remainder as $3^{35}$ when divided by $7$. 
I was stuck here, and I wanted to try to reverse engineer the answer. I know the remainder is $5$. Therefore I should be able to write $3^{35} -5 + 5$ as $7k+5$. However, I have no idea how to factor out a $7$ from $3^{35}-5$, or from $10^{35}-5.$ How could I find this $k$ non-explicitly?

Do you know Fermat little theorem?
for prime $p, n^p \equiv n \mod p\\
n^{p-1}\equiv 1 \mod p$
$10^35 = (10^6)^5(10^5)\\
10^35 \equiv (10^5) \mod 7\\
10^35 \equiv 5 \mod 7$
Alternative:
You could say, $10^{35} -1 = (10^7 - 1)(10^{28} + 10^{21} + 10^{14} + 10^7 + 1)$
And the remainder of the product equals the product of the remainders.
$(10^7 - 1)(10^{28} + 10^{21} + 10^{14} + 10^7 + 1) / 7\\
(3-1)(3^4 + 3^3 + 3^2 + 3 + 1) = 242$
and the remainder of 242/7 = 4
