{"provider_url":"https://hatena.blog","author_name":"hamayanhamayan","categories":["Security"],"blog_url":"https://blog.hamayanhamayan.com/","height":"190","author_url":"https://blog.hatena.ne.jp/hamayanhamayan/","title":"pingCTF 2025 Writeup","published":"2025-03-26 23:34:25","image_url":null,"url":"https://blog.hamayanhamayan.com/entry/2025/03/26/233425","provider_name":"Hatena Blog","description":"[crypto] easy rsa n\u306e\u7279\u5b9a RSA\u306e\u5f0f\u3092\u3046\u307e\u304f\u4f7f\u3063\u3066\u8a08\u7b97\u3092\u623b\u3059 \u30bd\u30eb\u30d0 [web] calc [web] sprint-user [web] keyboard-lovers [crypto] easy rsa Every ctf needs RSA challenge :) from Crypto.Util.number import getPrime, bytes_to_long import os p = getPrime(1024) q = getPrime(1024) n = p*q e = getPrime(512) d = pow(e, -1, (p-1)*(q-\u2026","type":"rich","width":"100%","blog_title":"\u306f\u307e\u3084\u3093\u306f\u307e\u3084\u3093\u306f\u307e\u3084\u3093","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Fblog.hamayanhamayan.com%2Fentry%2F2025%2F03%2F26%2F233425\" title=\"pingCTF 2025 Writeup - \u306f\u307e\u3084\u3093\u306f\u307e\u3084\u3093\u306f\u307e\u3084\u3093\" class=\"embed-card embed-blogcard\" scrolling=\"no\" frameborder=\"0\" style=\"display: block; width: 100%; height: 190px; max-width: 500px; margin: 10px 0px;\"></iframe>","version":"1.0"}