Understanding a question 
If c is randomly chosen from the integers 20 to 99, inclusive, what is
  the probability that $c^3-c$ is divisible by 12?

I have not got the question that what is implying by c? 

How to get the outcomes divisible by 12 ?

 A: Think to a table:
$$
\begin{array}{c|cc}
c & c^3-c \\
\hline
20 & 7980 & \bullet \\
21 & 9240 & \bullet \\
22 & 10626 \\
\vdots & \vdots \\
98 & 941094 \\
99 & 970200 & \bullet
\end{array}
$$
How many of the numbers in the right column are divisible by $12$ (marked with $\bullet$)? Let's say they are $d$; assuming that “random” means ”uniformly random”, you have $d$ favorable outcomes out of $80$, so the probability is $d/80$.
Now you have two paths: doing all the computations or finding a cleverer way that involves writing $c^3-c=(c-1)c(c+1)$ and reasoning about it.
A: HINT: (Edited to include egreg's comments)
No matter whatever $c$ you take $c^3-c = (c-1).c.(c+1)$ is divisible by $3! = 6$.
So for $c^3-c$ to be divisible by 12, you will need $(c-1).c.(c+1)$ as even.odd.even and not odd.even.odd except when this even in the later expression  is a multiple of 4.
So probability = ?
A: Hint: factorise $c^3-c$. Observe that the factors are consecutive integers with the middle one being $c$.
Now think about what constraints must apply to $c$ for divisibility by $12=3 × 4$. For example if $c$ is odd then the product of the other two integers is divisible by $4$ so $c$ needs to be an odd multiple of $3$ unless one of the others is an even multiple of $3$ and so forth.
