Let $f(k)=a-k^{2015}$. Because $(k-27)\mid f(k)$, we have $f(27)=0$, i.e. $27$ is a root of $f(k)$. This means that $a-27^{2015} = 0$ which implies that the last two digits of $a$ is the same as those of $27^{2015}$. $$27^{2015}=3^{6045} = (3^4)^{1511}\times 3 = 81^{1511} \times 3$$

Now by applying the "quick way to find the tens digit" trick, we know the tens digit of $81^{1511}$ is $8$. Meanwhile, its units digit obviously is $1$. Hence, $81^{1511}$ ends with $81$ which leads to $27^{2015}$ ends with $\boxed{43}$.