It is known that for any integer $K \neq 27$, the number $a-K^{3}$ is divisible by $27-K$. Find $a$.
Note that $\frac{a-K^{3}}{27-K}=\frac{a-27^{3}}{27-K}+\frac{27^{3}-K^{3}}{27-K}=\frac{a-27^{3}}{27-K}+27^{2}+27 K+K^{2}$. Therefore, the number $\frac{a-27^{3}}{27-K}-$

is an integer for any $K \neq 27$, which means the number

$a-27^{3}$ is divisible by any integer except zero, hence $a=27^{3}$.

## Answer

$a=3^{9}$.