CMIMC 2017

The operation given in the question basically finds the reciprocal of the sum of reciprocal of the numbers being operated upon, that is $x\otimes y=\dfrac{1}{\dfrac{1}{x}+\dfrac{1}{y}}$

Now operating from the innermost brackets of the problem we have: $2^{2016} \otimes 2^{2017}= \dfrac{2^{2016}.2^{2017}}{2^{2016}+2^{2017}}= \dfrac{2^{2017}}{3}=\dfrac{2^{2017}}{1+2}$

Further: $2^{2015} \otimes \dfrac{2^{2017}}{3}= \dfrac{2^{2015}.\dfrac{2^{2017}}{3}}{2^{2015}+\dfrac{2^{2017}}{3}}= \dfrac{2^{2017}}{7}= \dfrac{2^{2017}}{1+2+4}$

In general we can say that the $n^{th}$ term for the series of operations $2^{1}*\otimes (2^{2}\otimes (2^{3}\otimes...\otimes(2^{n-1}\otimes 2^{n})))$ would yield the result $\dfrac{2^{n}}{\displaystyle{\sum_{r=0}^{n-1} {2^{r}}} } = \dfrac{2^{n}}{2^{n}-1}$

Hence for the given problem : $\left \lfloor 2^{1} \otimes (2^{2}\otimes(2^{3}\otimes... \otimes(2^{2016}\otimes 2^{2017})))\right \rfloor=\left \lfloor \dfrac{2^{2017}}{2^{2017}-1} \right\rfloor= \left \lfloor \dfrac{1}{1-\dfrac{1}{2^{2017}}} \right \rfloor=\boxed{1}$

Note that $x \otimes y = \dfrac {xy}{x+y} = \dfrac 1{\frac 1x+\frac 1y}$ . Then we have:

$\begin{aligned} X & = 2^1 \otimes \left(2^2 \otimes \left(2^3 \otimes \cdots \otimes \left(2^{2016} \otimes 2^{2017} \right) \right) \right) \\ & = 2^1 \otimes \left(2^2 \otimes \left(2^3 \otimes \cdots \otimes \left(2^{2015} \otimes \left(\frac 1{\frac 1{2^{2016}} + \frac 1{2^{2017}}} \right) \right) \right)\right) \\ & = 2^1 \otimes \left(2^2 \otimes \left(2^3 \otimes \cdots \otimes \left(2^{2014} \otimes \left(\frac 1{\frac 1{2^{2015}} + \frac 1{2^{2016}} + \frac 1{2^{2017}}} \right) \right) \right)\right) \\ & = 2^1 \otimes \left(2^2 \otimes \left(2^3 \otimes \cdots \otimes \left(2^{2013} \otimes \left(\frac 1{\frac 1{2^{2014}} + \frac 1{2^{2015}} + \frac 1{2^{2016}} + \frac 1{2^{2017}}} \right) \right) \right)\right) \\ & = \ \cdots \\ & = \frac 1{\displaystyle \sum_{n=1}^{2017} \left(\frac 12\right)^n} = \frac 1{\frac 12 \left(\frac {1-\frac 1{2^{2017}}}{1-\frac 12} \right)} = \frac {2^{2017}}{2^{2017}-1} \end{aligned}$

$\implies \lfloor X \rfloor = \left \lfloor \dfrac {2^{2017}}{2^{2017}-1} \right \rfloor = \boxed{1}$