[GmatG]

there is a formula to calculate the total no. of factors of n :

n = x^a * y^b * z^c .. where x,yz are prime no. and a,b,c are +ve integers..
than, total no. of factors of n will be
(a+1)(b+1)(c+1).

Applying the above formula in the question.. n = 2^4 * 3 * 5

a=4, b=1 and c = 1

So, total no. of factors will be (4+1)(1+1)(1+1) ie, 20. --- (1)

factors of n are greater than or equal to 8 and less than or equal to 30factors of n are greater than or equal to 8 and less than or equal to 30

8<= factor <= 30
2^3 <= factor <= 2*3*5

now, finding no. factors less than 2^3
(a =0, b=0 and c=0) 1
(a =1, b=0 and c=0) 2
(a =2, b=0 and c=0) 4
(a =0, b=1 and c=0) 3
(a =0, b=0 and c=1) 5
(a =1, b=1 and c=0) 6

no. of fators less than 2^3 will be 6

no. of fators greater than 2*3*5
(a =2, b=1 and c=1) 60
(a =3, b=1 and c=1) 120
(a =4, b=1 and c=1) 240
(a =4, b=1 and c=0) 48
(a =4, b=0 and c=1) 80
(a =3, b=0 and c=1) 40
will be 6

Total no. of factors that does not fall into the range of
2^3 <= factor <= 2*3*5 will be 12 (6+6) ---- (2)

hence required no. of factors = 8 (1) - (2)

Hope that helps. Thanks