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Xophe92/rsa.py

Last active October 8, 2019 19:29
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b00tl3gRSA2 - Points: 400 - (Solves: 1076)
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rsa.py
import binascii
n = 62452183886922897079565462589532968595884552052214229613445442774743423725102737845742975666636854617775795947653844346559950723662263034642750426162273832839101527738528294467648817519626232830075222929548053450371355548382455233416933866954353433351778542716327032301670100608617444861845487978703599835379
c = 56347572031538334377661578217905779910490772992642826875089038058436862705266996954941611050094273478907514997960956816289297147965363014381184730266460034420695156137130424878759263215224108315562552968913202468760795267651517029708252934757877155640216359411075081754078898540094203891997354635495949480953
e = 61897578533882795669670180546914170241934801387789726178209999472564491631659190026553473372696292044152567645449981835158789679959142094011319619930275962936609685337352059111535711982420122060517738442734450137974442442074077044188542576386785694404751980458638816026283776944151705012059689090902604045873
for d in range(65537,65537): # 0, 100000000
if pow(pow(c, d, n), e, n) == c:
print(d)
print(pow(c, d, n))
if d%100 == 0:
print(".", flush=True, end="")
d = 65537
print(pow(pow(c, d, n), e, n) == c)
m = pow(c, d, n)
print(binascii.unhexlify(hex(m)[2:]))
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