Sharif University CTF 2016 - Smooth As Silk

Category: Crypto Points: 200 Solves: 11 Description:

 

p > q

n = p*q = 11461532818525251775869994369956628325437478510485272730419843320517940601808597552925629451795611990372651568770181872282590309894187815091191649914833274805419432525386234876477307472337026966745408507004000948112191973777090407558162018558945596691322909404307420094584606103206490426551745910268138393804043641876816598599064856358266650339178617149833032309379858059729179857419751093138295863034844253827963

flag = md5(str(p))

EN:

Can solve it by using Pollard P-1 Factorization Method（http://www.mersennewiki.org/index.php/P-1_Factorization_Method）;

step 01:select B1

N =p*q (p>q)

q<=sqrt(N)

B1=sqrt(N)

more B1 is bigger, the more possible can find out it(?! maybe, No!)

 

step 02: count out E2, E3,E5 ....

2^E2<=B1,  3^E3<=B1 ...

E2=log(B1)/log(2)

E3=log(B1)/log(3)

E5=log(B1)/log(5)

E7=log(B1)/log(7)

...

push 2  in the z[] E2 times, push 3 in the z[] E3 times, push 5 in the z[] E5 times ...

 

step 03:

x=a, (a is a prime)

i=0

do

　　x^z[i]≡a(mod n)

　　x=a

until  gcd(n, x-1) !=1

gcd(n,x-1) is one factor of N.

 

#!/usr/bin/python3
import math

n=11461532818525251775869994369956628325437478510485272730419843320517940601808597552925629451795611990372651568770181872282590309894187815091191649914833274805419432525386234876477307472337026966745408507004000948112191973777090407558162018558945596691322909404307420094584606103206490426551745910268138393804043641876816598599064856358266650339178617149833032309379858059729179857419751093138295863034844253827963
z=[]
prime=[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997];
def gcd(a,b):
    if b==0:
        return a
    return gcd(b,a%b)

def e(a,b):
    return pow(a,b)%n

def mysqrt(n):
    x=n
    y=[]
    while(x>0):
        y.append(x%100)
        x=x//100
    y.reverse()
    a=0
    x=0
    for p in y:
        for b in range(9,-1,-1):
            if(((20*a+b)*b)<=(x*100+p)):
                x=x*100+p - ((20*a+b)*b)
                a=a*10+b
                break

    return a

B1=mysqrt(n)
for j in range(0,len(prime)):
    for i in range(1, int(math.log(B1)/math.log(prime[j]))+1):
        z.append(prime[j])

#print(z)

for pp in prime:
    i=0
    x=pp
    while(1):
        x=e(x,z[i])
        i=i+1
        y=gcd(n,x-1)
        if(y!=1):
            print (y)
            exit(0)
        if(i>=len(z)):
            break

 

python3 sss.py, we can the number:

958483424448747472504060861580795018746355733561446016442794600533395417361061386707061258449029078376132360127073305093209304646989718030495000998517698501250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

 

then

n=11461532818525251775869994369956628325437478510485272730419843320517940601808597552925629451795611990372651568770181872282590309894187815091191649914833274805419432525386234876477307472337026966745408507004000948112191973777090407558162018558945596691322909404307420094584606103206490426551745910268138393804043641876816598599064856358266650339178617149833032309379858059729179857419751093138295863034844253827963

p=958483424448747472504060861580795018746355733561446016442794600533395417361061386707061258449029078376132360127073305093209304646989718030495000998517698501250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

q=11957987510443514049047696785587234758227153373363589891876816598599064856358266650339178617149833032309379858059729179857419751093138295863034844253827963

flag=md5(str(p)) = c78504a558bdb6213b9019f6925fa4ae

flag is c78504a558bdb6213b9019f6925fa4ae

 

CN:

这个是因子分解，用 Pollard P-1因子分解法（http://www.mersennewiki.org/index.php/P-1_Factorization_Method）, pyhton3 源码看上面。

还是要看B1的选择，选择不好也是有可能解不出，解不出就重新选择，直到解出。