SOLUTION: How many factors of 3•6•9•12•15•18 are greater than 1 and are the square of an integer? Options:- 15 14 7 6 Answer is 14 but how?? Thanks for any help.

 3 =     3
 6 = 2∙  3
 9 =     3∙3
12 = 2∙2∙3
15 =     3  ∙5
18 = 2∙  3∙3

So 3∙6∙9∙12∙15∙18 = 24∙38∙5 

A perfect square must have an even power
of each of its prime factors.  Since we want
factors of  24∙38∙5 the exponents of
the primes must not exceed the exponents of
2,3,and 5 in 24∙38∙5. 

All factors of 3•6•9•12•15•18 are of the form

2x∙3y∙5z, where x, y, and z are even integers,
[including 0].

The possible even powers of 2 are 0, 2, and 4.

Therefore there are 3 ways to choose the power of 2. 

The possible powers of 3 are 0,2,4,6 and 8.

So there are 5 ways to choose the power of 3.

The only possible power of 5 is 0.

So there is only 1 way to choose the power of 5, (as 0)

The answer would be 3 ways times 5 ways times 1 way 

= 3∙5∙1 = 15 ways.

However, since it says "greater than 1" we must subtract 
1 from the 15, so 

Answer = 14.

Here are all 15 factors of 3∙6∙9∙12*15∙18:

 1. 203050 = 1 <--can't use this!
 2. 203250 = 9
 3. 203450 = 81
 4. 203650 = 729
 5. 203850 = 6561
 6. 223050 = 4
 7. 223250 = 36
 8. 223450 = 324
 9. 223650 = 2916
10. 223850 = 26244
11. 243050 = 16
12. 243250 = 144
13. 243450 = 1296
14. 243650 = 11664
15. 243850 = 104976

Edwin