Plzz find all the 2-digit numbers each of which has n divisors for a)n=8 b)n=9

We need to find 2 digit numbers (10-99) each of which has 
a) 8 divisors
b) 9 divisors
 
Here's what I found:
For 8 divisors, the number must have
    a. 3 distinct prime factors, or
    b. of the form {p^3, q}
    which are:
2^3*3=24 (1,2,3,4,6,8,12,24)
2*3*5=30 (1,2,3,5,6,10,15,30)
2^3*5=40 (1,2,4,5,8,10,20,40)
2*3*7=42 (1,2,3,6,7,14,21,42)
2*3^3=54 (1,2,3,6,9,18,27,54)
2^3*7=56 (1,2,4,7,8,14,28,56)
2*3*11=66 (1,2,3,6,11,22,33,66)
2^3*11=88 (1,2,4,8,11,22,44,88)
2*3*13=78 (1,2,3,6,13,26,39,78)
2*5*7=70 (1,2,5,7,10,14,35,70)

For 9 divisors, it is necessary that the 2-digit number be a perfect square, and the square-root is composite.
For this, we can list only
(2*3)^2=36  (1,2,3,4,6,9,12,18,36)

So there are 10 2-digit numbers with 8 factors, and 1 with 9 factors.

Note: looks like 72 has 12, the most factors in two digit numbers:
72 (1,2,3,4,6.8,9,12,18,24,36,72)

Number nine is f(x)= 40x : ) hoped i ya