Problem - 4253
Show that there exists an infinite number of integers in the form of $(2^n+27)$ which are multiples of $7$.
Note that $8\equiv 1\pmod{7}$. Therefore $2^{3k}\equiv 1\pmod{7}$ holds for any positive integer $k$. It follows that $$2^{3k} + 27\equiv 1 + 27\equiv 0\pmod{7}$$
This means when $n$ is a multiple of $3$, the number $(2^n+27)$ will be a multiple of $7$. Clearly, there is an infinite number of such $n$.