If x^3 is a positive factor of 10!*11!, how many possible integer values of x are there?
[10! x 11!] ==144850083840000 = 2^16 * 3^8 * 5^4 * 7^2 * 11
x==36 possible integer values as follows:
x ==1 , 2 , 3 , 4 , 5 , 6 , 8 , 9 , 10 , 12 , 15 , 16 , 18 , 20 , 24 , 30 , 32 , 36 , 40 , 45 , 48 , 60 , 72 , 80 , 90 , 96 , 120 , 144 , 160 , 180 , 240 , 288 , 360 , 480 , 720 , 1440 , Total == 36 such integers.
