<PRE><B>Question 1200580</B>
&lt;br&gt;
By whatever means you want, determine that the prime factorization of 222264 is&lt;br&gt;
(2^3)(3^4)(7^3)&lt;br&gt;
So any factor of 222264 is of the form&lt;br&gt;
(2^a)(3^b)(7^3)&lt;br&gt;
where 
0&lt;=a&lt;=3
0&lt;=b&lt;=4
0&lt;=c&lt;=3&lt;br&gt;
If the factor is a perfect square, then a, b, and c all have to be even.  So&lt;br&gt;
a can be 0 or 2  (2 choices)
b can be 0, 2, or 4  (3 choices)
c can be 0 or 2  (2 choices)&lt;br&gt;
By the fundamental counting principle, the number of factors of 222264 that are perfect squares is 2*3*2 = 12.&lt;br&gt;
ANSWER: 12&lt;br&gt;
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