Question 1062628
<pre><b><font size=5>
 3 =     3
 6 = 2&#8729;  3
 9 =     3&#8729;3
12 = 2&#8729;2&#8729;3
15 =     3  &#8729;5
18 = 2&#8729;  3&#8729;3

So 3&#8729;6&#8729;9&#8729;12&#8729;15&#8729;18 = 2<sup>4</sup>&#8729;3<sup>8</sup>&#8729;5 

A perfect square must have an even power
of each of its prime factors.  Since we want
factors of  2<sup>4</sup>&#8729;3<sup>8</sup>&#8729;5 the exponents of
the primes must not exceed the exponents of
2,3,and 5 in 2<sup>4</sup>&#8729;3<sup>8</sup>&#8729;5. 

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

2<sup>x</sup>&#8729;3<sup>y</sup>&#8729;5<sup>z</sup>, 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&#8729;5&#8729;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&#8729;6&#8729;9&#8729;12*15&#8729;18:

 1. 2<sup>0</sup>3<sup>0</sup>5<sup>0</sup> = 1 <--can't use this!
 2. 2<sup>0</sup>3<sup>2</sup>5<sup>0</sup> = 9
 3. 2<sup>0</sup>3<sup>4</sup>5<sup>0</sup> = 81
 4. 2<sup>0</sup>3<sup>6</sup>5<sup>0</sup> = 729
 5. 2<sup>0</sup>3<sup>8</sup>5<sup>0</sup> = 6561
 6. 2<sup>2</sup>3<sup>0</sup>5<sup>0</sup> = 4
 7. 2<sup>2</sup>3<sup>2</sup>5<sup>0</sup> = 36
 8. 2<sup>2</sup>3<sup>4</sup>5<sup>0</sup> = 324
 9. 2<sup>2</sup>3<sup>6</sup>5<sup>0</sup> = 2916
10. 2<sup>2</sup>3<sup>8</sup>5<sup>0</sup> = 26244
11. 2<sup>4</sup>3<sup>0</sup>5<sup>0</sup> = 16
12. 2<sup>4</sup>3<sup>2</sup>5<sup>0</sup> = 144
13. 2<sup>4</sup>3<sup>4</sup>5<sup>0</sup> = 1296
14. 2<sup>4</sup>3<sup>6</sup>5<sup>0</sup> = 11664
15. 2<sup>4</sup>3<sup>8</sup>5<sup>0</sup> = 104976

Edwin</pre></b></font>