# EasyCTF_2018: Hidden Key

Category: Cryptography Points: 250 Description:

Ugh, another RSA problem? Help me decrypt this message please file.

## Write-up

This challenge was way too simple for a 250 point challenge. From the start we are given 4 variables, n, e, c and 2d+phi(n). Well, there's a problem right there. d is derived from e^-1 mod(phi(n)) and 2d is just 2e^-1 mod(phi(n)). If you add phi(n) to any equation with a mod(phi(n)), it would simply get modulo-ed and wrap around having no effect. As such, our 2d+phi(n) is just 2d. To get d, just divide by 2.

# 2d+phi(n)=37732317402097892656548210368059150520715311244186827321460515339110825509994606476441219328574059681654107935415953369600005791889525955066919184173730927777542439132137456187316767857542892181069605145389028008447003177489965519989581906591202747002126347010093645064766054828212106925552838747800759084251589866089273026254966234164176466563186438248776961541146620434009503689100128122320997739428424975306256417865610162074912031638292878093662018512614000114412433772628503843778614582386663028723327141014781803346402661832265806675785945746391801332346416280561063520342062390151626459813123281193179325458386
s = 37732317402097892656548210368059150520715311244186827321460515339110825509994606476441219328574059681654107935415953369600005791889525955066919184173730927777542439132137456187316767857542892181069605145389028008447003177489965519989581906591202747002126347010093645064766054828212106925552838747800759084251589866089273026254966234164176466563186438248776961541146620434009503689100128122320997739428424975306256417865610162074912031638292878093662018512614000114412433772628503843778614582386663028723327141014781803346402661832265806675785945746391801332346416280561063520342062390151626459813123281193179325458386

# Calculate d
d = s // 2


Now we can just decrypt the message using Python.

# ./hidden-key.py
easyctf{1n1n77n9x2db57dqbk}


Therefore, the flag is easyctf{1n1n77n9x2db57dqbk}.