RSA: How to calculate the private exponent? - Cryptography Stack Exchange most recent 30 from crypto.stackexchange.com 2022-01-20T21:15:18Z https://crypto.stackexchange.com/feeds/question/25547 https://creativecommons.org/licenses/by-sa/4.0/rdf https://crypto.stackexchange.com/q/25547 0 RSA: How to calculate the private exponent? free https://crypto.stackexchange.com/users/24370 2015-05-09T16:19:59Z 2015-05-09T16:49:34Z <p>I have this problem: In RSA algorithm considering $n=33$ (modulus) and public exponent $e=3$, calculate the corresponding private exponent $d$.</p> <p>I know that $d = e^{-1} \pmod{\varphi(n)}$ and $\varphi(n) = (p – 1) (q – 1)$ but I don't know $p$ and $q$.</p> <p>How can I do this?</p> https://crypto.stackexchange.com/questions/25547/-/25548#25548 1 Answer by SEJPM for RSA: How to calculate the private exponent? SEJPM https://crypto.stackexchange.com/users/23623 2015-05-09T16:26:47Z 2015-05-09T16:31:58Z <p>If I'm understanding your question right, you want to obtain $d$ from given $n$ and $e$.</p> <p>You'll have to factor $n=33=3*11$ and as $N=p*q$ you have obtained your $p=11$ and $q=3$. Now proceed as usual with calculating the inverse.</p> <p>As pointed out correctly above, you can't easily generalize this approach to larger numbers as factoring $n$ will be infeasible. Now you see how to factor the number but for larger numbers even the best algorithms fail.<br> Concerning your $e=3$ I'd consider using Coppersmith's attacks on low exponent RSA.</p>