Fixed Hamming distance property of binary deletion correcting codes

Question

Let $x=(x_1x_2...x_n)$ be a binary sequence of length $n$. The Varshamov-Tenengolts code $VT_0(n)$ consists of all binary vectors $(x_1, . . . , x_n)$ satisfying $\Sigma_{i=1}^n i*x_i \equiv0 \pmod {n+1} $.

Prove that $\forall$ $x,y \in VT_0(n)$ which has equal hamming weight the Hamming distance between $x$ and $y$ is exactly 4. For a binary vectore $x$ the hamming weight is $w$ if $\Sigma_{i=1}^n x_i= w $.

Answer 1

Take $n=11$. 11000000100 and 00111000000 are both in the code and they both have Hamming weight 3 but the Hamming distance between them is 6.