Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 15C / d / iv / Solution

Here $N^2=441$, so the least power of two exceeding it is $Q=2^9=512$. The continued fraction iswhose convergents beginThe only nontrivial convergent with denominator below $N=21$ is $5/6$, and indeedThe uniqueness result therefore givesAs a check, $10^6\equiv 1(\mathrm{mod}21)$, while $10^2\not\equiv 1$ and $10^3\not\equiv 1$, so the order is exactly six.

Solved by gpt-5.6-sol high.