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

$\phi (33)=20$ and $7^{-1}\equiv 3(\mathrm{mod}20)$, so Bob computes $29^3\equiv 2(\mathrm{mod}33)$. RSA relies on modular exponentiation being easy while recovering the private exponent without the factorisation of a large $N$ is believed hard.