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Dec 11: Crypt-o-Math 2.0
Challenge
You bruteforced last years math lessons? This time you cant escape!
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c = (a * b) % p
c=0x423EDCDCDCD928DD43EAEEBFE210E694303C695C20F42A27F10284215E90
p=0xB1FF12FF85A3E45F722B01BF3135ED70A552251030B114B422E390471633
b=0x88589F79D4129AB83923722E4FB6DD5E20C88FDD283AE5724F6A3697DD97
find a to get your flag.
Solution
All we need to do is solve the modular equation, the divm function in gmpy2 does exactly that:
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import codecs
import gmpy2
def int_to_text(number):
return codecs.decode(format(number, 'x'), 'hex').decode('utf-8')
# c = (a * b) % p
c = 0x559C8077EE6C7990AF727955B744425D3CC2D4D7D0E46F015C8958B34783
p = 0x9451A6D9C114898235148F1BC7AA32901DCAE445BC3C08BA6325968F92DB
b = 0xCDB5E946CB9913616FA257418590EBCACB76FD4840FA90DE0FA78F095873
# solve the equation
a = gmpy2.divm(c, b, p)
print(int_to_text(a))
Flag
HV17-XtDw-0DzO-YRgB-2b2e-UWNz