惯性聚合 高效追踪和阅读你感兴趣的博客、新闻、科技资讯
阅读原文 在惯性聚合中打开

推荐订阅源

IT之家
IT之家
Microsoft Azure Blog
Microsoft Azure Blog
人人都是产品经理
人人都是产品经理
博客园 - 聂微东
博客园_首页
阮一峰的网络日志
阮一峰的网络日志
V
V2EX
小众软件
小众软件
F
Fortinet All Blogs
Microsoft Security Blog
Microsoft Security Blog
奇客Solidot–传递最新科技情报
奇客Solidot–传递最新科技情报
H
Hackread – Cybersecurity News, Data Breaches, AI and More
量子位
Google DeepMind News
Google DeepMind News
Jina AI
Jina AI
让小产品的独立变现更简单 - ezindie.com
让小产品的独立变现更简单 - ezindie.com
aimingoo的专栏
aimingoo的专栏
B
Blog RSS Feed
Cyber Security Advisories - MS-ISAC
Cyber Security Advisories - MS-ISAC
宝玉的分享
宝玉的分享
有赞技术团队
有赞技术团队
J
Java Code Geeks
WordPress大学
WordPress大学
The Cloudflare Blog

Homepage on Aditya Telange

One Year with evil-winrm-py - A Retrospective Bypassing LinkedIn's Connection Privacy with a Simple Search Filter Making Dynamic Instrumentation Accessible with Frida UI Breaking Payload Encryption in Web Applications HackTheBox (HTB) - Escape HackTheBox (HTB) - Resolute HackTheBox (HTB) - Certified State of VMWare Workstation (Pro?) on Linux Android App Security Testing Lab with MobSleuth Android phone as a Webcam on Linux Breaking down Reverse shell commands HackTheBox (HTB) - Photobomb Merging AOSP Security Patches into Custom ROMs Primer on HTTP Security Headers Image Zoom-In effect with HUGO HackTheBox (HTB) - Legacy HackTheBox (HTB) - Lame Cryptohack - Keyed Permutations [5 pts] Cryptohack - Resisting Bruteforce [10 pts] Cryptohack - Base64 [10 pts] Cryptohack - Bytes and Big Integers [10 pts] Cryptohack - Hex [5 pts] Cryptohack- XOR Starter [10 pts] HackTheBox (HTB) - Horizontall HackTheBox (HTB) - Forge HackTheBox (HTB) - Previse HackTheBox (HTB) - BountyHunter HackTheBox (HTB) - Explore HackTheBox (HTB) - Cap HackTheBox (HTB) - Pit
Cryptohack - RSA Starter 1 [10 pts]
[Aditya Telange](https://x.com/adityatelange) · 2022-05-20 · via Homepage on Aditya Telange

The Solution is shared considering CAN I SHARE MY SOLUTIONS?

Problem

All operations in RSA involve modular exponentiation.

Modular exponentiation is an operation that is used extensively in cryptography and is normally written like: 210 mod 17

You can think of this as raising some number to a certain power (210 = 1024), and then taking the remainder of the division by some other number (1024 mod 17 = 4). In Python there’s a built-in operator for performing this operation: pow(base, exponent, modulus)

In RSA, modular exponentiation, together with the problem of prime factorisation, helps us to build a “trapdoor function”. This is a function that is easy to compute in one direction, but hard to do in reverse unless you have the right information. It allows us to encrypt a message, and only the person with the key can perform the inverse operation to decrypt it.

Find the solution to 10117 mod 22663

Solution

Python3

FLAG := 19906