BUU-Crypto-[HDCTF2019]basic rsa

[HDCTF2019]basic rsa

Title Description ： Got flag Please wrap it flag{} Submit .
The problem solving steps ： Open the attachment python file

import gmpy2
from Crypto.Util.number import *
from binascii import a2b_hex, b2a_hex

flag = "*****************"

p = 262248800182277040650192055439906580479
q = 262854994239322828547925595487519915551

e = 65533
n = p * q

c = pow(int(b2a_hex(flag), 16), e, n)

print c
# 27565231154623519221597938803435789010285480123476977081867877272451638645710

The title gives p,q,e,c,c Namely # The number after ：
27565231154623519221597938803435789010285480123476977081867877272451638645710 了 , Equivalent to known p,q,e,c,n, seek M that will do
exp as follows ：

import gmpy2
import libnum

p = 262248800182277040650192055439906580479
q = 262854994239322828547925595487519915551

e = 65533
n = p * q

c = 27565231154623519221597938803435789010285480123476977081867877272451638645710
phi = (p - 1) * (q - 1)
d = gmpy2.invert(e, phi)
M = pow(c, d, n)
print libnum.n2s(M)

Run to get results flag

flag：flag{B4by_Rs4}