finding numbers to make both sides equal 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?

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).
