<PRE><B>Question 1039490</B>
&lt;pre&gt;&lt;b&gt;
If p is a prime factor of a perfect square,
then the maximum k for which p&lt;sup&gt;k&lt;/sup&gt; is a factor,
then k MUST be an even number.

We prime factor 2352 with a factor tree:

          2352
          /  \
         2  1176 
            /  \
           2   588
               / \
              2  294
                 / \
                2  147
                  /  \
                 3   49
                    /  \
                   7    7

So 2352 = 2&lt;sup&gt;4&lt;/sup&gt;&amp;#8729;3&amp;#8729;7&lt;sup&gt;2&lt;/sup&gt;

The primes 2 and 7 already have even powers,
so we must multiply by 3 so that the 3 will
also have an even power.

So we must multiply by 3:

2352&amp;#8729;3 = 2&lt;sup&gt;4&lt;/sup&gt;&amp;#8729;3&lt;sup&gt;2&lt;/sup&gt;&amp;#8729;7&lt;sup&gt;2&lt;/sup&gt;

The answer is 3.  Then the perfect square will be
7056, which is 2&lt;sup&gt;4&lt;/sup&gt;&amp;#8729;3&lt;sup&gt;2&lt;/sup&gt;&amp;#8729;7&lt;sup&gt;2&lt;/sup&gt; 
and (2&lt;sup&gt;2&lt;/sup&gt;&amp;#8729;3&amp;#8729;7)&lt;sup&gt;2&lt;/sup&gt; = (4&amp;#8729;3&amp;#8729;7)&lt;sup&gt;2&lt;/sup&gt; = 84&lt;sup&gt;2&lt;/sup&gt;

Edwin&lt;/pre&gt;&lt;/B&gt;</PRE>