jmc

We are looking for an integer $a$ such that $27a$ is congruent to 1 modulo 28. In other words, we want to solve $$27a\equiv 1(\mathrm{mod}28).$$We subtract $28a$ from the left-hand side to obtain $-a\equiv 1(\mathrm{mod}28)$. This congruence is equivalent to the previous one since $28a$ is a multiple of 28. Next we multiply both sides by $-1$ to obtain $a\equiv -1(\mathrm{mod}28)$. Thus $28-1=\boxed{27}$ is the modular inverse of 27 (mod 28). (Note that since $(m-1{)}^2=m^2-2m+1\equiv 1(\mathrm{mod}m)$, we always have that $m-1$ is its own inverse modulo $m$.)