Here is almost a similar question. But in that case, $3-\sqrt{5}$ is a proper fraction (i.e. $0<3-\sqrt{5}<1$). So, if we expand both $(3-\sqrt{5})^{34}$ and $(3+\sqrt{5})^{34}$ using binomial theorm and do some calculations (including cancellations), we will reach to a conclusion.
My problem which is different. It should be solved without using calculators. Average solving time for each problem in the exam (that includes this problem) is $3$ minutes:
What is the remainder when dividing $\left \lfloor (6+\sqrt{7})^8 \right \rfloor$ by $9$?
$\text{(A) 0}$
$\text{(B) 1}$
$\text{(C) 2}$
$\text{(D) 4}$
$\text{(E) 8}$
Any help/hint would be appreciated.
It's not actually an answer, but it's to long for a comment. I am sorry. I tried to use the binomial theorem, but it seems to be very hard. Here is, what i tried.
We have \begin{align*} (6+\sqrt 7)^8&=\sum_{2\mid k}\binom{8}{k}6^{8-k}7^{k/2} +\sqrt 7\cdot \sum_{2\nmid k}\binom{8}{k}6^{8-k}7^{(k-1)/2} \\ &= \sum_{k=0}^4\binom{8}{2k}6^{8-2k}\cdot 7^{k}+\sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k\\ &=7^4+\underbrace{6^2\cdot\sum_{k=0}^3\binom{8}{2k}6^{6-2k}\cdot 7^{k}}_{\text{divisible by $9$}}+\sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k \end{align*} One can calulate with much patience, that $$ \lfloor(6+\sqrt 7)^8\rfloor\equiv (-2)^4+\left\lfloor \sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k\right\rfloor\equiv -2+\left\lfloor \sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k\right\rfloor\equiv -2 + \lfloor 5896848 \sqrt 7\rfloor.$$ Maybe, you can get until here without a calculator. But now I stuck. Sorry sorry