TITLE: A Gift Problem for the Year 2018
QUESTION [23 upvotes]: We had this problem in exam  class  yesterday on Combinatoric  and it was supposed to be the new year gift from our teacher. The exercise was entitled A Gift Problem for the Year 2018 

Problem:
The numbers  $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots,\frac{1}{2018} $ are written on the blackboards. John chooses any two numbers say $x$ and $y$ erases them and writes the number $x+y+xy$.
  He continues to do so until there is only one number left on the board.
  What are the possible value of the final number?

I understood the problem as follows for instance if John take $x=1$ and $y=\frac{1}{2}$ then $x+y+xy =2$ and the new list becomes $$2,\frac{1}{3},\frac{1}{4},\cdots,\frac{1}{2018} $$ 
 continuing like this and so on.....
Please bear with me  that I do not want to propose my solution since I fell like it was wrong and I don't want to fail the exam before the result get out. but by the way I found, $2017$, $2018$ and $2019$ but I am still suspicious. 
You may help is you have an idea.

REPLY [39 votes]: Consider the multiplicative  law on $\Bbb R$ defines by $$x*y =x+y+xy =(x+1)(y+1)-1 $$
you can check that it is associative and commutative on $\Bbb R$. Therefore at the end the remaining number is 
$$\begin{align}x_0*x_1*x_2*\cdots x_{2018} &= 1*\frac{1}{2}*\frac{1}{3}*\cdots *\frac{1}{2018} \\&=\left[\prod_{i=1}^{2018}(1+x_i)\right]-1\\
&=\left[\prod_{i=1}^{2018}\left(1+\frac{1}{i}\right)\right]-1 \\
&=\frac{2}{1}\cdot \frac{3}{2}\cdot \frac{4}{3}\cdot \ldots \cdot \frac{2018+1}{2018}-1=\color{red}{2019-1=2018.}
\end{align}$$