If a and b are positive integers, and (2^3)(3^4)(5^7)=(a^3)∗b, how many different possible values of b are there?
Pick.
A) 2
B) 3
C) 4
D) 6
E) 12
Let's assume 2^3,3^3 and 5^3 as three different entities.   The entity 5^3 is there twice.   Thus, out of these three, we can take none, 1, two or all three. This gives a total of :   3C0+3C1+3C2+3C3 = 1+3+3+1= 
<gadget id="calculator">1+3+3+1</gadget>
<output>8</output>
8 ways. Now the other 5^3 will not make a difference when being selected in the mentioned way. It will only make a difference when it is either selected with5^3 only(1 case), or when both of them together are considered with either 2^3(1 case) or 3^3(1 case). Finally, when all the 4 are considered together, we have one final case. Thus a total of   8+3+1= 
<gadget id="calculator">  8+3+1</gadget>
<output>12</output>
12 cases.   ANSW12R:12
<result>E</result>
If a and b are positive integers, and (2^3)(3^4)(5^7)=(a^3)∗b, how many different possible values of b are there?