SMIC (1)

intro crypto RSA FCSC 2020 solved on

Description

RSA encryption relies on the modular exponentiation of large integers. Using standard notations, compute the “ciphertext” c corresponding to the following plaintext m: m = 29092715682136811148741896992216382887663205723233009270907036164616385404410946789697601633832261873953783070225717396137755866976801871184236363551686364362312702985660271388900637527644505521559662128091418418029535347788018938016105431888876506254626085450904980887492319714444847439547681555866496873380 using the public key: (n, e) = (115835143529011985466946897371659768942707075251385995517214050122410566973563965811168663559614636580713282451012293945169200873869218782362296940822448735543079113463384249819134147369806470560382457164633045830912243978622870542174381898756721599280783431283777436949655777218920351233463535926738440504017, 65537).

This flag is FCSC{xxxx}, where xxxx has to be replaced by the value of c written in decimal.

A variant of this challenge is available here: SMIC (2).

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DateAuthor Language Tags Vote
2023-12-02
whoo
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2023-12-04
BuzzYbis
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2024-07-30
GarlicHorse
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