$\varphi(33)=20$ and $7^{-1}\equiv3\pmod {20}$, so Bob computes $29^3\equiv2\pmod {33}$. RSA relies on modular exponentiation being easy while recovering the private exponent without the factorisation of a large $N$ is believed hard.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 2D / b / Solution

 Solution (source code)