Find the greatest integer in the expression below as a function of the given conditions

Question

Find the greatest integer less than $\sqrt{2^{100}+10^{10}}.$
Answer: $2^{50}$

I tried, but I can't finish:

$\sqrt{2^{100}+(2\cdot5)^{10}}=\sqrt{2^{10}(2^{90}+5^{10})}=2^5\sqrt{2^{90}+5^{10}}$

Answer 1

A little GM-AM can also be helpful:

\begin{eqnarray*} \sqrt{2^{100}+10^{10}} & = & 2^{50} \sqrt{ 1+\frac{10^{10}}{2^{100}}} \\ & \stackrel{GM-AM}{<} & 2^{50}\frac{1+1+\frac{10^{10}}{2^{100}}}2 \\ & = & 2^{50} + \frac{10^{10}}{2^{51}} \\ & < & 2^{50} + \frac{10^{10}}{2^{50}} \\ & = & 2^{50} + \left( \frac{10}{2^5}\right)^{10} \\ & < & 2^{50} + 1 \end{eqnarray*}

Answer 2

Hint: Can you show that $\left(2^{50}\right)^2<2^{100}+10^{10}<\left(2^{50}+1\right)^2$?

Answer 3

It results that

$2^{50}=\sqrt{2^{100}}<\sqrt{2^{100}+10^{10}}<\sqrt{2^{100}+16^{10}}=$

$=\sqrt{2^{100}+2^{40}}<\sqrt{2^{100}+2^{51}+1}=\sqrt{\left(2^{50}+1\right)^2}=2^{50}+1$

Hence ,

$2^{50}<\sqrt{2^{100}+{10^{10}}}<2^{50}+1\;.$

Consequently,

$\left\lfloor\sqrt{2^{100}+10^{10}}\right\rfloor=2^{50}\;.$