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This question already has an answer here:

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.

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marked as duplicate by Peter Taylor, user1729, ahulpke, Strants, David Hill Feb 1 '18 at 21:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Please show your work even if it is not correct. That way we can help and find any errors. $\endgroup$ – robjohn Jan 23 '18 at 16:35
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    $\begingroup$ Since we are not in your class, how about an actual title? $\endgroup$ – Asaf Karagila Jan 23 '18 at 18:51
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    $\begingroup$ Something more descriptive of the actual mathematical problem? $\endgroup$ – Asaf Karagila Jan 23 '18 at 20:16
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    $\begingroup$ Okay, then therefore it's my downvote until you find a better title. $\endgroup$ – Asaf Karagila Jan 23 '18 at 20:21
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    $\begingroup$ What's the excuse for rejecting the title suggested by @AccidentalFourierTransform, is it not click baity enough? $\endgroup$ – Asaf Karagila Jan 24 '18 at 6:05
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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 $$\begin{align}x_0*x_1*x_2*\cdots x_{2018} &= 1*\frac{1}{2}*\frac{1}{3}*\cdots *\frac{1}{2018} \\&=\left[\prod_{i=1}^{2018}(1+x_i)\right]-1\\ &=\left[\prod_{i=1}^{2018}\left(1+\frac{1}{i}\right)\right]-1 \\ &=\frac{2}{1}\cdot \frac{3}{2}\cdot \frac{4}{3}\cdot \ldots \cdot \frac{2018+1}{2018}-1=\color{red}{2019-1=2018.} \end{align}$$

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  • $\begingroup$ This works with any number, btw. Not just 2018. $\endgroup$ – PyRulez Jan 24 '18 at 5:09
  • $\begingroup$ you are right it actually a telescopic product. there is no reason to restrict it to 2018 $\endgroup$ – user518372 Jan 24 '18 at 7:38
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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?

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  • $\begingroup$ It's 2018, @JohnLou You have to subtract the one again. $\endgroup$ – Thomas Andrews Jan 23 '18 at 15:38
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    $\begingroup$ Is $(1+1)(1/2+1)\cdots(1/2018+1)$ computed by rewriting things as $(2)(3/2)(4/3)\cdots(2019/2018)$ and recognizing that it telescopes? $\endgroup$ – Mauve Jan 23 '18 at 15:44
  • $\begingroup$ Thanks @ThomasAndrews $\endgroup$ – John Lou Jan 23 '18 at 15:48

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