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Why everytime the final number comes the same?

Suppose we write the integers 1 thru $n$, choose 2 random ones, erase them, and replace them with the single integer that is their sum plus their product instead. We now have $n-1$ integers written. We repeat this process until we only have 1 number written. Prove that there is only one result possible.

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Aliya
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  • What does this have to do with probability? – Chris Eagle Jul 04 '12 at 18:35
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    @ChrisEagle You choose random integers? ;) Seriously though, this should be retagged, but I'm not sure what to. – tomasz Jul 04 '12 at 18:37
  • I’ve retagged it (abstract-algebra) on the basis of my solution, but if anyone can find a better tag, feel free to change it. I’ve also added (puzzle). – Brian M. Scott Jul 04 '12 at 18:41
  • It is a random process that may have a deterministic outcome. – Michael R. Chernick Jul 04 '12 at 18:42
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    This is a cool problem, but the original poster doesn't give any context or their own take on it. – Ben Millwood Jul 04 '12 at 18:43
  • I would keep both tags. It does have a proabilistic aspect to it. – Michael R. Chernick Jul 04 '12 at 18:43
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    @Michael: The problem as stated has no probabilistic aspect: that is in fact the point of it. The fact that there is a related probabilistic problem isn’t relevant, and the (probability) tag would be misleading. – Brian M. Scott Jul 04 '12 at 18:49
  • @BrianM.Scott The probability tag is not misleading. The problem as posed is probabilist and can be approached that way. Problems can involve probabilistic calculations that lead to degenerate distributions and the fact that a problem has an algebraic solution does not make it an algebra problem although I think the other tags are appropriate. – Michael R. Chernick Jul 04 '12 at 19:22
  • @Michael: I'd say if there is no probability distribution specified in the question, then it's probably(?) not a probability question. Otherwise any problem involving arbitrary choices becomes a probability problem. –  Jul 04 '12 at 19:29
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    @Michael: I flatly disagree. The probability tag is misleading for anyone using tags to see whether his question has already been answered or to look for answers to similar questions. – Brian M. Scott Jul 04 '12 at 19:30
  • @RahulNarain There is a probability distribution specified in the question. Let me remind you by quoting from it. "Suppose we write the integers 1 thru n, choose 2 random ones,". So there is a probability distribution in the question. It is the distribution for pairs of integers selected at random out of the integers 1 to n. Based on this one can determine the distribution for the sum + product term computed at each time the board is erased. This distribution is non-degenerated at every stage except the last and the solution can be determined probabilistically. – Michael R. Chernick Jul 04 '12 at 19:54
  • The final distribution happens to be deterministic but that does not change it from being a probbility problem. You can tag it as an algebra problem if you like since there is a purely algebraic solution. But the fact that one method of solution is not probabilistic does not change it from being a probability problem. – Michael R. Chernick Jul 04 '12 at 19:57
  • @BrianMScott How can you say the probability tag is misleading? It is clearly a probability problem regardless of whether or not it can be solved by other means. – Michael R. Chernick Jul 04 '12 at 19:59

2 Answers2

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For any numbers $a$ and $b$, $a+b+ab=(a+1)(b+1)-1$; call this $a\otimes b$. For any $a,b,c$,

$$\begin{align*} (a\otimes b)\otimes c&=\Big((a+1)(b+1)-1\Big)\otimes c\\ &=(a+1)(b+1)(c+1)-1\\ &=a\otimes\Big((b+1)(c+1)-1\Big)\\ &=a\otimes(b\otimes c)\, \end{align*}$$

and clearly $a\otimes b=b\otimes a$, so $\otimes$ is a commutative, associative binary operation. It follows that the final result is simply $$\bigotimes_{k=1}^nk=\prod_{k=1}^n(k+1)-1=(n+1)!-1$$ irrespective of the order in which the calculations are performed.

Brian M. Scott
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$$L = \left(\displaystyle \prod_{k=1}^{n} (1+x_k) \right)- 1 = \left(\displaystyle \prod_{k=1}^{n-2} (1+x_k) \right) \left(1+x_{n-1} \right) \left(1+x_n \right)- 1\\ = \left(\displaystyle \prod_{k=1}^{n-2} (1+x_k) \right) \left(1+\left(x_{n-1} + x_n + x_{n-1}x_n \right) \right)- 1 = \left(\displaystyle \prod_{k=1}^{n-2} (1+x_k) \right) \left(1+\tilde{x}_{n-1} \right)- 1$$ Hence, $L$ remains same, if we replace $(x_{n-1},x_n) \to (x_{n-1} + x_n + x_{n-1} x_n)$