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Ten people are sitting in a row, and each is thinking of a negative integer no smaller than $-15$. Each person subtracts, from his own number, the number of the person sitting to his right (the rightmost person does nothing). Because he has nothing else to do, the rightmost person observes that all the differences were positive. Let $x$ be the greatest integer owned by one of the 10 people at the beginning. What is the minimum possible value of $x$?

I was completely thrown off this problem because of the sentence, "Let $x$ be the greatest integer owned by one of the 10 people at the beginning.", which I intepretted as the first value we start with is -1, because the largest negative integer is -1. It doesn't even come off as ambigious, and yet, to find the minumum value of x, the first value needs to be -15. So, I'm curious at what I'm missing here. Word problems have never been my forté, so any advice on how one should approach these types of questions would be appreciated. Thanks!

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  • $\begingroup$ What do you mean by "first value" ? $\endgroup$ – Ted Sep 15 '17 at 5:13
  • $\begingroup$ The first value of x that we start with, which would be the rightmost person. $\endgroup$ – Swanty Sep 15 '17 at 5:14
  • $\begingroup$ Okay. The rightmost person cannot have value -1. From the conditions of the problem, the values have to decrease as you go from left to right. So the rightmost person has to have the smallest value. $\endgroup$ – Ted Sep 15 '17 at 5:19
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The question, simplified:

There are ten integers $-15\le x_1,\dots,x_{10}\le-1$ with $x_1-x_2,x_2-x_3,\dots,x_9-x_{10}>0$. What is the smallest possible value of $\max x_i$?

Immediately we see $x_1>x_2>x_3>\dots>x_{10}\ge-15$, that $\max x_i=x_1$ and that the minimum gap between adjacent $x_i$ is one. The minimum $x$ turns out to be $-6$: $$(-6,-7,-8,-9,-10,-11,-12,-13,-14,-15)$$

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