Primary Knowledge

Catégorie: Crypto Difficulté: very-easy Flag: HTB{0h_d4mn_4ny7h1ng_r41s3d_t0_0_1s_1!!!}

Challenge

1KB

crypto_primary_knowledge.zip

archive

Description

---

Surrounded by an untamed forest and the serene waters of the Primus river, your sole objective is surviving for 24 hours. Yet, survival is far from guaranteed as the area is full of Rattlesnakes, Spiders and Alligators and the weather fluctuates unpredictably, shifting from scorching heat to torrential downpours with each passing hour. Threat is compounded by the existence of a virtual circle which shrinks every minute that passes. Anything caught beyond its bounds, is consumed by flames, leaving only ashes in its wake. As the time sleeps away, you need to prioritise your actions secure your surviving tools. Every decision becomes a matter of life and death. Will you focus on securing a shelter to sleep, protect yourself against the dangers of the wilderness, or seek out means of navigating the Primus’ waters?

Analyse du code

Il s’agit d’une implémentation basique de RSA sauf que N est un nombre premier

Donc on peut calculer ϕ(N) très facilement puisque ϕ(N) = N - 1

Grâce à cela on peut calculer l’inverse de e dans N qui sera notre exposant d pour déchiffrer le message.

Pour les explications détaillées du fonctionnement de RSA, je renvoie vers Wikipédia a minima

---

Script de résolution

from Crypto.Util.number import long_to_bytes

def solve():
	n = 144595784022187052238125262458232959109987136704231245881870735843030914418780422519197073054193003090872912033596512666042758783502695953159051463566278382720140120749528617388336646147072604310690631290350467553484062369903150007357049541933018919332888376075574412714397536728967816658337874664379646535347
	e = 65537
	c = 15114190905253542247495696649766224943647565245575793033722173362381895081574269185793855569028304967185492350704248662115269163914175084627211079781200695659317523835901228170250632843476020488370822347715086086989906717932813405479321939826364601353394090531331666739056025477042690259429336665430591623215

	# Calcul de l'inverse de e dans ϕ(n)
	d = pow(e, -1, n-1)
	# Déchiffrement du message
	m = pow(c, d, n)
	flag = long_to_bytes(m).decode()
	print(f"Flag: {flag}")

if __name__ == '__main__':
	solve()