5lb book, Advance Quant question #7

[mohit.wrangler]

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

Here is an excerpt from answer's explanation (5lb book of GRE practice problems, page 883, question #7):

In any set of three consecutive integers, a multiple of 3 will be included. Thus, (c-1)c(c+1) is always divisible by 3 for any integer c. This takes care of part of the 12. So the question becomes "How many of the possible (c-1)c(c+1) values are divisible by 4?"

I agree that (c-1)c(c+1) is always divisible by 3 for any integer c. But then the question becomes "How many of the possible (c-1)c(c+1)/3 values are divisible by 4?" i.e. how can we ignore the fact that 4 has to divide quotient which will result after division by 3.

[tommywallach]

The question "Which of these numbers is divisible by 12?" is actually the same question as "Which of these numbers is divisible by both 3 and 4?" because any number divisible by both 3 and 4 is divisible by 12.

So the point of the explanation is that we ALREADY know every number in that function will be divisible by 3, so all we have to worry about is the 4 part. Make sense?

-t