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  <resource>
  <id>7160</id>
  <path>/www/nrich/html/content/id/7160/</path>
  <resourceTypeID>1</resourceTypeID>
  <last_published>2011-02-01T00:00:01</last_published>
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&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;p&gt;Shakil wants to remove numbers from the set $\{1,2,3,..., 16\}$ so that no two remaining numbers add to make a perfect square. What is the smallest number of numbers that he needs to remove?&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
If you liked this problem, &lt;a href=&quot;http://nrich.maths.org/790&quot;&gt;here is an NRICH task&lt;/a&gt; which challenges you to use similar mathematical ideas.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
  &lt;/p&gt;&lt;/mdoxml&gt;</indexXML>
  <solutionXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
The following seven pairs add to $16$ so at least one of each pair
must be removed:&lt;br&gt;&lt;/br&gt;
$(1,15)$, $(2,14)$, $(3,13)$, $(4,12)$, $(5,11)$, $(6,10)$,
$(7,9)$.&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
If removing these seven is sufficient, then we would be left with
$8$, $16$ and seven others.&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
But&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;margin-left: 40px;&quot;&gt;  $16+9 = 25$    
 So we must remove $9$    and keep its partner
$7$&lt;/div&gt;

&lt;div style=&quot;margin-left: 40px;&quot;&gt;    $7+2 = 9$  
     So we must remove $2$     and
keep $14$&lt;/div&gt;

&lt;div style=&quot;margin-left: 40px;&quot;&gt;$14+11 = 25$      So we
must remove $11$   and keep $5$&lt;/div&gt;

&lt;div style=&quot;margin-left: 40px;&quot;&gt;    $5+4 = 9$    
   So we must remove $4$     and keep
$12$&lt;/div&gt;

&lt;div style=&quot;margin-left: 40px;&quot;&gt;$12+13 = 25$      So we
must remove $13$   and keep $3$&lt;/div&gt;

&lt;div style=&quot;margin-left: 40px;&quot;&gt;    $3+1 = 4$    
   So we must remove $1$     and keep
$15$&lt;/div&gt;

&lt;div style=&quot;margin-left: 40px;&quot;&gt;$15+10 =
25$      So we must remove $10$   and
keep $6$&lt;/div&gt;

&lt;div&gt; &lt;/div&gt;

&lt;div&gt;But we have kept $3$ and
$6$ which add to $9$.&lt;/div&gt;

&lt;div&gt; &lt;/div&gt;

&lt;div&gt;Hence it is not
sufficient to remove only seven. If we remove the number $6$, we
obtain a set which satisfies the condition:
$\{8,16,7,14,5,12,3,15\}$ or in ascending order
$\{3,5,7,8,12,14,15,16\}$.&lt;/div&gt;

&lt;div&gt; &lt;/div&gt;

&lt;div&gt;Hence eight is the
smallest number of numbers that may be removed.&lt;/div&gt;

&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</solutionXML>
  <noteXML/>
  <clueXML/>
  <canonXML/>
  <end_user_role>2</end_user_role>
  <difficulty>3</difficulty>
  <keystage1>0</keystage1>
  <keystage2>0</keystage2>
  <keystage3>1</keystage3>
  <keystage4>1</keystage4>
  <keystage4plus>0</keystage4plus>
  <title>Weekly Problem 12 - 2011</title>
  <description>Weekly Problem 12 - 2011</description>
  <spec_group>Numbers and the Number System
    <specifier>Square numbers</specifier>
  </spec_group>
  <spec_group>Using, Applying and Reasoning about Mathematics
    <specifier>Working systematically</specifier>
  </spec_group>
  <spec_group>Calculations and Numerical Methods
    <specifier>Addition &amp; subtraction</specifier>
  </spec_group>
</resource>