According to the given information $A = 2^n \cdot 10^m + 14^n$, where $10^{m-1} < 14^n < 10^m$. Assume $n$ is even i.e. $n=2k$. Then $A \equiv 14^n \equiv (-1)^{2k} = 1 \pmod{5}$, i.e. $5 \mid A-1$ Next assume $n$ is odd, i.e. $n=2k-1$. Then $A \equiv (-1)^{2k+1} \cdot 1^m + (-1)^{2k+1} = - 1 - 1 = -2 \pmod{3}$, i.e. $3 \mid A - 1$. Hence $A-1$ is either divisible by 3 or 5. Consequently (since $A>5$) $A-1$ cannot be a prime.