A Gift Problem for the Year 2018

Question

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.

Answer 1

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 $$\begin{align}x_0*x_1*x_2*\cdots x_{2018} &= 1*\frac{1}{2}*\frac{1}{3}*\cdots *\frac{1}{2018} \\&=\left[\prod_{i=1}^{2018}(1+x_i)\right]-1\\ &=\left[\prod_{i=1}^{2018}\left(1+\frac{1}{i}\right)\right]-1 \\ &=\frac{2}{1}\cdot \frac{3}{2}\cdot \frac{4}{3}\cdot \ldots \cdot \frac{2018+1}{2018}-1=\color{red}{2019-1=2018.} \end{align}$$

Answer 2

Hint: Note that $xy+x+y = (x+1)(y+1) - 1$. In particular, this implies that at any stage of the process, if all the numbers on the board are $\{a_1, \cdots, a_k\}$, then the product $$(a_1 + 1) (a_2 + 1) \cdots (a_k + 1)$$ is invariant, i.e. it doesn't change when you erase two numbers $x$, $y$, and replace them with $xy+x+y$. What does this imply the final number must be?