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?
 A: By Fermat's little theorem, $3^6\equiv 1 \bmod 7$, so that $3^{35}\equiv 3^5\equiv 5\bmod 7$. Hence $7\mid (3^{35}-5)$, and explicitly, $3^{35}=7\cdot 7147363585571386.$
Edit: "shifting interest to the factorisation": $$3^{35}-5=2\cdot 7\cdot 1729363\cdot 2066472911,$$ using integer factorisation algorithms.
A: If you insist on algebraic techniques then note $\ 3(3^{35}-5)\, =\, (3^{36}-1)-14$
and $\,\ 7 = \color{#c00}{3^2\!-\!3\!+\!1}\mid 3^6\!-\!1\mid 3^{36}-1\ $ so $\,7\,$ divides both summands above, so $\,7\mid 3^{35}\!-\!5$
where $\ \color{#c00}{x^2-x+1}\mid x^6-1\ $ since it divides the $\,\rm\color{#c00}{difference\ of\ squares}\,$ below
$$ x^6-1\ =\, (x^2-1)\,(\!\overbrace{x^4+x^2+1}^{\Large\color{#c00}{ (x^2+1)^2-x^2}}\!)$$
Remark $\ $ "Algebraic" factors that arise arise from polynomial factorizations are heavily employed when factoring integers of the form $\rm\:b^n\pm 1.\:$ A good place to learn about such is Wagstaff's splendid introduction to the Cunningham Project, whose goal is to factor numbers of the form $\rm\:b^n\pm 1.\:$ There you will find  mentioned not only old results such as Legendre (primitive divisors of $\rm\:b^n\pm 1\:$ are $\rm\,\equiv 1\pmod{2n},$ but also newer results, e.g. those exploiting cyclotomic factorizations. e.g. see below.

Often number identities are more perceptively viewed as special cases of function or polynomial identities. For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\  D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations (e.g. see below).
$$\begin{array}{rl}
x^4 + 2^2 \quad=&  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\[0.5em]
\dfrac{x^6 + 3^2}{x^2 + 3} \quad=&  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\[0.5em]
\dfrac{x^{10} - 5^5}{x^2 - 5} \quad=&  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\[0.5em]
\dfrac{x^{12} + 6^6}{x^4 + 36} \quad=&  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36)
\end{array}$$
A: You can use the fact that the remainder of $ab$ is the remaineder of $a$ multiplied by the remaineder of $b$. So forgeting factors of $7$, we have $3^{35}=27*3^{32}=6*(9)^{16}=6*2^{16}=6*(16)^4=6*2^4=6*2=5$
A: 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
A: 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 $$3^{35}-5=3^{33}3^2-5=(7N-1)\times 9-5 = 7\times 9N-9-5=7\times (9N-2)$$
