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  <author_name>jeneshicc</author_name>
  <author_url>https://blog.hatena.ne.jp/jeneshicc/</author_url>
  <blog_title>落書き、時々落学</blog_title>
  <blog_url>https://jeneshicc.hatenadiary.org/</blog_url>
  <categories>
    <anon>Project Euler</anon>
  </categories>
  <description>Consider all integer combinations of ab for 2 a 5 and 2 b 5: 22=4, 23=8, 24=16, 25=32 32=9, 33=27, 34=81, 35=243 42=16, 43=64, 44=256, 45=1024 52=25, 53=125, 54=625, 55=3125If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:4, 8, …</description>
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  <provider_name>Hatena Blog</provider_name>
  <provider_url>https://hatena.blog</provider_url>
  <published>2008-10-29 00:15:49</published>
  <title>Problem 29</title>
  <type>rich</type>
  <url>https://jeneshicc.hatenadiary.org/entry/20081029/1225293349</url>
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