Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$.
Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$.
Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$.
Then, if any absolute value equals 1, then $a_n=0$.
Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria.
Assume that to be the only way the criteria is met.
To prove, let $|y-x|>1$, and $|z-y|>1$. Then, $a_4 \ge 2z$, $a_5 \ge 4z$, and $a_6 \ge 4z$.
However, since the minimum values of $a_5$ and $a_6$ are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be $z=1$, $|y-x|=2$. Again assume that any other scenario will not meet criteria.
To prove, divide the other scenarios into two cases: $z>1$, $|y-x|>1$, and $|z-y|>1$; and $z=1$, $|y-x|>2$, and $|z-y|>1$.
For the first one, $a_4 \ge 2z$, $a_5 \ge 4z$, $a_6 \ge 8z$, and $a_7 \ge 16z$, by which point we see that this function diverges.
For the second one, $a_4 \ge 3$, $a_5 \ge 6$, $a_6 \ge 18$, and $a_7 \ge 54$, by which point we see that this function diverges.

Therefore, the only scenarios where $a_n=0$ is when any of the following are met:
$|y-x|<2$ (280 options)
$|z-y|<2$ (280 options, 80 of which coincide with option 1)
$z=1$, $|y-x|=2$. (16 options, 2 of which coincide with either option 1 or option 2)
Adding the total number of such ordered triples yields $280+280-80+16-2=\boxed{494}$.

Note to author:
Because $a_4 \ge 2z$, $a_5 \ge 4z$, $a_6 \ge 8z$, and $a_7 \ge 16z$ doesn't mean the function diverges. What if $z = 7$, $a_4 = 60$, and $a_5 = 30$ too?

Note to the Note to author:
This isn't possible because the difference between is either 0, 1, or some number greater than 1. If $a_4 = 63$ (since it must be a multiple of $z$), then a_5 is either 0, 63, or some number greater than 63.