{"provider_url":"https://hatena.blog","type":"rich","blog_url":"https://jeneshicc.hatenadiary.org/","version":"1.0","blog_title":"\u843d\u66f8\u304d\u3001\u6642\u3005\u843d\u5b66","height":"190","provider_name":"Hatena Blog","width":"100%","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Fjeneshicc.hatenadiary.org%2Fentry%2F20081029%2F1225293349\" title=\"Problem 29 - \u843d\u66f8\u304d\u3001\u6642\u3005\u843d\u5b66\" class=\"embed-card embed-blogcard\" scrolling=\"no\" frameborder=\"0\" style=\"display: block; width: 100%; height: 190px; max-width: 500px; margin: 10px 0px;\"></iframe>","published":"2008-10-29 00:15:49","title":"Problem 29","author_name":"jeneshicc","url":"https://jeneshicc.hatenadiary.org/entry/20081029/1225293349","author_url":"https://blog.hatena.ne.jp/jeneshicc/","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, \u2026","image_url":null,"categories":["Project Euler"]}