Problem Solving (PS) | Re: If n is a positive two-digit integer, how many different values of n a

n^3-n= n(n^2-1)= n(n-1)(n-2)-> 3 consecutive integers
3 consecutive integers are divisible by 3!=6. Thus 3 is a factor. For the product to be divisible by 12, it should also be divisible by 4.
Considering 2 cases for the value of n:
n is odd->
Then n+1 and n-1 will be divisible by 2, thus the product will be divisible by 4.
Half the numbers between 10 and 99 [90 numbers: (99-10)+1] are odd -> 45 numbers

n is even->
Then n-1 and n+1 will be odd. The product will be divisible by 4 only when n is divisible by 4. The number of numbers divible by 4 between 10 and 99 is 22

Add 45 and 22= 67 numbers