<PRE><B>Question 803431</B>
&lt;pre&gt;
n^2 = 10*k+r  0 &lt;u&gt;&lt;&lt;/u&gt; r &lt;u&gt;&lt;&lt;/u&gt; 9

r is the units (last) digit of a perfect square
Since 

0²=0
1²=1, 
2²=4, 
3²=9
4²=16
5²=25
6²=36
7²=81
8²=64
9²=81

the last (units) digit of ALL perfect squares can
only be 0,1,4,5,6, or 9.  Thus r can be
0,1,4,5,6, or 9, r cannot be 2,3,7, or 8.
 
0&lt;sup&gt;2&lt;/sup&gt; = 10(0) + 0
1&lt;sup&gt;2&lt;/sup&gt; = 10(0) + 1
2&lt;sup&gt;2&lt;/sup&gt; = 10(0) + 4
3&lt;sup&gt;2&lt;/sup&gt; = 10(0) + 9
4&lt;sup&gt;2&lt;/sup&gt; = 10(1) + 6
5&lt;sup&gt;2&lt;/sup&gt; = 10(2) + 5
6&lt;sup&gt;2&lt;/sup&gt; = 10(3) + 6
7&lt;sup&gt;2&lt;/sup&gt; = 10(4) + 9

Edwin&lt;/pre&gt;
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