What is $\lfloor(\sqrt 3 +1)^5\rfloor$? (without a calculator)

Question

The question states:

Let $[x]$ be the greatest integer less than or equal to $x$. If $x=(\sqrt 3 +1)^5$, then $[x]$ is equal to

1. $75$
2. $50$
3. $152$
4. $151$

When I punch it out in the calculator, I get $x \approx 152.210235533$. Therefore, the correct choice is 3. But the problem is how do I find it without using a calculator, should such problems appear in competitive exams?

Answer 1

Consider $N=(1+\sqrt 3)^5+(1-\sqrt 3)^5$. This is an integer.

Since $-1\lt 1-\sqrt 3\lt 0$, we have $$0\lt -(1-\sqrt 3)^5\lt 1.$$

So, we have $$\begin{align}\lfloor (1+\sqrt 3)^5\rfloor&=\lfloor N-(1-\sqrt 3)^5\rfloor\\&=N\\&=(1+\sqrt 3)^5+(1-\sqrt 3)^5\\&=2\left((\sqrt 3)^0\binom 50+(\sqrt 3)^2\binom 52+(\sqrt 3)^4\binom 54\right)\\&=152\end{align}$$