I just needed some help for this problem they gave us at our mathematics class, its’s an Olympiad type question, I suspect the answer is 2019, I tried with smaller cases and tried to use a recursive relation, or some type of induction. I also tried factoring by noticing that $a+b+ab=(a+1)(b+1)-1$ but couldn’t manage to solve it. Hoping someone could help. Here is the problem:
Martin wrote the following list of numbers on a whiteboard:
$$1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5},…,\frac{1}{2019}$$
Martin asked for a volunteer. Vincent offered to play, Martin explained the rules to Vincent. Vincent has to choose two of the numbers on the board, lets say $a$ and $b$. He has to wipe off these numbers and write the number $a+b+ab$ on the board.
For example, if Vincent would have chosen $\frac{1}{2}$ and $\frac{1}{3}$, he would wipe these numbers off and write $\frac{1}{2}+\frac{1}{3}+\frac{1}{2}·\frac{1}{3}=1$ on the board. Then the board would look like this:
$$1, 1, \frac{1}{4}, \frac{1}{5},…,\frac{1}{2019}$$
Martin asked Vincent to do this over and over again choosing any two numbers each time, until only one number was left. When Vincent was done, Martin asked him to open an envelope where he had written a prediction, Vincent opened it and was surprised to find that the number on the whiteboard and on the envelope were the same.
What was Martin’s prediction? Justify your answer.
Best Answer
Let the terms on the board in any given move be $a_1, a_2, a_3, a_4 ... a_n$. We will prove that $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ never changes, and therefore is an invariant.
First, consider any move, with $a_i$ and $a_j$. Notice that $(1+a_i)(1+a_j) = 1+a_i+a_j+a_ia_j$, and therefore the value of $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ constant.
Let the value that Martin gets at the end be $n$. Notice that $(1+\frac{1}{1})(1+\frac{1}{2})...(1+\frac{1}{2019})-1 = \frac{2}{1}\frac{3}{2}....\frac{2020}{2019} - 1 = (1 + n) - 1 \implies n = 2019$, so we are done.