Question 30470
 X*Y*Z=4000 = 4*(2*5)^3 = 2^5* 5^3.

 so X, Y, Z are in the form 2^i * 5^j  (X= 2^i1 * 5^j1,..
 with i1+i2+i3 = 5 and j1+j2+j3 = 3.

 there are C(3+5-1, 5) = C(7,5) = 21 possible i's(why?)
 and C(3+3-1, 3) = C(5,3) = 10 possible j's

 Totally, there are 21*10 = 21 possible ordered triples (X,Y,Z).

 Try X*Y = 2^3* 5^2

 Kenny