Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 12I / ii / Solution

Sum all nonzero vectors of ${\mathbb{F}}_2^d$. In each coordinate, exactly $2^{d-1}$ vectors have entry one, which is even because $d>1$. The sum is therefore zero, soThus the all-one word belongs to the Hamming code and has Hamming weight $n$.