Set $k=11^a \cdot k'$ and $\ell=11^b\cdot \ell'$ where $11\nmid k'\ell'$. It suffices to prove $k'=\ell'$. Suppose the contrary that $k'\ne \ell'$, let $k'>\ell'$ wlog. Take $m\equiv \textstyle\frac{1}{11}(k'+1)\pmod{k'}$, where such an $m$ exists since $(k',11)=1$. Then, $(11m-1,k)=(11m-1,k') = k'$. At the same time, $(11m-1,\ell)=(11m-1,\ell')\mid \ell'<k'$, so $(11m-1,\ell)<(11m-1,k)$.