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?

As a (possibly) different approach:  start by remarking that $$3^3=4\times 7-1$$
It follows that $$3^{33}=(4\times 7-1)^{11}=7N-1$$
Where $N$ is a simple expression in powers of $4$ and $7$ (with some binomial coefficients thrown in for good measure). Specifically  $$N=\sum_{i=1}^{11}\binom {11}i (-1)^{11-i}4^i7^{i-1}$$
We then remark that $$3^{35}-5=3^{33}3^2-5=(7N-1)\times 9-5 = 7\times 9N-9-5=7\times (9N-2)$$
