# Challenge Name: opisthocomus-hoazin
![date](https://img.shields.io/badge/date-15.06.2021-brightgreen.svg)  ![solved in time of CTF](https://img.shields.io/badge/solved-in%20time%20of%20CTF-brightgreen.svg)   ![crypto category](https://img.shields.io/badge/category-Cryptography-blueviolet.svg)   ![value](https://img.shields.io/badge/value-300-blue.svg)  
## Description
The plural of calculus is calculi.   - [opisthocomus-hoazin.py](opisthocomus-hoazin.py)- [ouput.txt](output.txt)
## Detailed solution
As we can see in the code the output print the value of n,e and ct  
```n = 15888457769674642859708800597310299725338251830976423740469342107745469667544014118426981955901595652146093596535042454720088489883832573612094938281276141337632202496209218136026441342435018861975571842724577501821204305185018320446993699281538507826943542962060000957702417455609633977888711896513101590291125131953317446916178315755142103529251195112400643488422928729091341969985567240235775120515891920824933965514217511971572242643456664322913133669621953247121022723513660621629349743664178128863766441389213302642916070154272811871674136669061719947615578346412919910075334517952880722801011983182804339339643e = 65537ct = [65639, 65645, 65632, 65638, 65658, 65653, 65609, 65584, 65650, 65630, 65640, 65634, 65586, 65630, 65634, 65651, 65586, 65589, 65644, 65630, 65640, 65588, 65630, 65618, 65646, 65630, 65607, 65651, 65646, 65627, 65586, 65647, 65630, 65640, 65571, 65612, 65630, 65649, 65651, 65586, 65653, 65621, 65656, 65630, 65618, 65652, 65651, 65636, 65630, 65640, 65621, 65574, 65650, 65630, 65589, 65634, 65653, 65652, 65632, 65584, 65645, 65656, 65630, 65635, 65586, 65647, 65605, 65640, 65647, 65606, 65630, 65644, 65624, 65630, 65588, 65649, 65585, 65614, 65647, 65660]```  We need to find the flag characters that give us (ord(flag_chars)^e)%n equal to ct values  
A python script that decrypt ct to get the flag
```pythonimport string
n = 15888457769674642859708800597310299725338251830976423740469342107745469667544014118426981955901595652146093596535042454720088489883832573612094938281276141337632202496209218136026441342435018861975571842724577501821204305185018320446993699281538507826943542962060000957702417455609633977888711896513101590291125131953317446916178315755142103529251195112400643488422928729091341969985567240235775120515891920824933965514217511971572242643456664322913133669621953247121022723513660621629349743664178128863766441389213302642916070154272811871674136669061719947615578346412919910075334517952880722801011983182804339339643e = 65537ciphertext = [65639, 65645, 65632, 65638, 65658, 65653, 65609, 65584, 65650, 65630, 65640, 65634, 65586, 65630, 65634, 65651, 65586, 65589, 65644, 65630, 65640, 65588, 65630, 65618, 65646, 65630, 65607, 65651, 65646, 65627, 65586, 65647, 65630, 65640, 65571, 65612, 65630, 65649, 65651, 65586, 65653, 65621, 65656, 65630, 65618, 65652, 65651, 65636, 65630, 65640, 65621, 65574, 65650, 65630, 65589, 65634, 65653, 65652, 65632, 65584, 65645, 65656, 65630, 65635, 65586, 65647, 65605, 65640, 65647, 65606, 65630, 65644, 65624, 65630, 65588, 65649, 65585, 65614, 65647, 65660]flag = ""
for i in ciphertext:     for j in string.printable:         if (ord(j)^e)%n == i:             flag+= j print(flag)```
## Flag
```flag{tH1s_ic3_cr34m_i5_So_FroZ3n_i"M_pr3tTy_Sure_iT's_4ctua1ly_b3nDinG_mY_5p0On}```