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Call a triple-x number an integer $k$ such that $k=x(x+1)(x+2)$ where $x \in Z$. How many triple-x numbers are there between 0 and 100,000?

I thought by doing $8!$ and $9!$ would work to see how many combinations there would be. I am not sure how to solve this problem. Can someone show me how to solve this?

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  • $\begingroup$ Either by solving $1000=x(x+1)(x+2)$ or by brute-force, you'll find out that $45$ is the smallest $x$ that works. So, anything below $45$ is fair game. $\endgroup$ – funktor Oct 28 '14 at 2:36
  • $\begingroup$ Have you tried to figure out how many integers $x$ satisfy $0\lt x\lt100000$? I believe $1$ is the smallest $x$ value that works. Not sure about the biggest, somewhere around the cube root of $100000$ I guess. $\endgroup$ – bof Oct 28 '14 at 2:39
  • $\begingroup$ @bof Sorry! I meant the largest. Silly error! $\endgroup$ – funktor Oct 28 '14 at 2:42
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First, let's note that for $x=-3$ or lower the product will be negative and thus, there aren't any values lower than this. $x=-2,-1,0$ all give the same value of 0. Note that $46*47*48=103776$ while $45*46*47=97290$ which would give 45 as the highest bound and 0 is the lowest, thus producing 46 numbers assuming that the ends are inclusive.

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Well, there's one with $x = 0$, one with $x = 1$, and keep going until you get something bigger than $100,000$. Equivalently, you're solving the inequality $x(x+1)(x+2) < 100,000$, which can be done in various ways (though there's no really clean way of doing it).

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