How to show that the triangle inequality holds for this metric? Define $d:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{R}$ by $\displaystyle d(m,n)=\frac{1}{\sup\{l\in\mathbb{N}: l!\text{ divides }\lvert m-n\rvert\}}$ with the obvious interpretation that when the supremum doesn't exist we define $d(m,n)=0$.
I'm having a bit of difficulty trying to show that the triangle inequality holds. Can someone give me some intuition on what direction I should head towards?
Thoughts: To show $d(m,n)\le d(m,k)+d(k,n)$, if one of either $\lvert m-k\rvert$ or $\lvert k-n\rvert$ odd then we're done. So we only need to consider the case where all of them are even. But I don't quite know how this helps.  
Also as a bit of an aside, how is this metric useful? It was just an example given in the notes and the details were left as an exercise.
 A: I don't think that you need to treat odd and even cases separately, but the parity observations are good ones. 
Assume that $m, k$ and $n$ are distinct integers, since the other cases are easily dealt with. Suppose that $l_1!$ is the maximal factorial dividing $|m-k|$ and $l_2!$ is the maximal factorial dividing $|k-n|$. Without loss of generality assume $l_1\leq l_2$. Note that this implies $l_1!$ divides $l_2!$, and so $l_1!$ divides $m-k+k-n=m-n$. If $l$ is the maximal natural number whose factorial divides $m-n$, it follows that $l$ is at least $l_1$ and: 
$$d(m,n) \leq \frac{1}{l_1} \leq \frac{1}{l_1}+\frac{1}{l_2} = d(m,k)+d(k,n).$$
I've never encountered this metric before, and unfortunately I haven't the foggiest idea how it might be "useful" (beyond providing an interesting exercise for those learning about metrics). Exploring properties of the topology associated to the metric might provide insight into what the metric is measuring, and this might suggest where it could arise naturally or be put to good use. Perhaps someone else will have a better idea on this part -- or you could ask the person who constructed those notes.
