Problem Solving (PS) | If an integer n is to be chosen at random from the integers

Here's my take, please correct me if I were wrong.

multiply out n(n+1)(n+2) = n^3 + 3n^2 + 2n.

Now, let's factor out n^3 from above equation which become n^3 (1+ 3n^-1 + 2n^-2) <-- I know they looks ugly but wait

You basically ignore (1+ 3n^-1 + 2n^-2) now so that you will know that to have n^3 divisible by 8; n must be the multiple of 2.

Here you will know that 96 / 2 = 48 numbers that are the multiple of 2.

Be aware here because 48 number have included the multiple of 8's but we still
...

---