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  <id>7153</id>
  <path>/www/nrich/html/content/id/7153/</path>
  <resourceTypeID>1</resourceTypeID>
  <last_published>2011-02-01T00:00:01</last_published>
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&lt;p&gt;Let $a= 2^{25}$, $b= 8^8$ and $c= 3^{11}$.&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
Now order the letters so that we get  ${?}&amp;amp;lt; {?}&amp;amp;lt; {?}$&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/6401&amp;amp;part=&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;
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By comparing &lt;span style=&quot;font-style: italic;&quot;&gt;b&lt;/span&gt; and &lt;span style=&quot;font-style: italic;&quot;&gt;c&lt;/span&gt; first, we have $b= 8^8=
(2^3)^8=2^{24}=(2^2)^{12}=4^{12}&amp;gt; 3^{11} = c$&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
so $c&amp;lt; b$.&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
But also $b= 2^{24}&amp;lt; 2^{25}=a$ so $b&amp;lt; a$.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Together these give $c&amp;lt; b&amp;lt; a$.&lt;br&gt;&lt;/br&gt;
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 &lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</solutionXML>
  <noteXML/>
  <clueXML/>
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  <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 6 - 2011</title>
  <description>Weekly Problem 6 - 2011</description>
  <spec_group>Numbers and the Number System
    <specifier>Powers &amp; roots</specifier>
  </spec_group>
  <spec_group>Algebra
    <specifier>Inequality/inequalities</specifier>
  </spec_group>
  <spec_group>Algebra
    <specifier>Using symbols</specifier>
  </spec_group>
</resource>