Solution. Among the divisors of a funny number, there cannot be a two or a prime $d$ with a remainder of 1 when divided by 3 (otherwise, the prime $d+2$ would be divisible by 3, implying $d=1$). Therefore, the divisors can only be threes (and their powers) and at most one prime divisor $p \equiv 2(\bmod 3)$ (if there is another such divisor $q$, then $d=p q \equiv 1(\bmod 3)$ and $d+2$, divisible by $3,$ is not prime). Moreover, the power of three cannot be greater than 4, since otherwise there would be a divisor $d=3^{5}$, for which $d+2=245$ is not prime.

Can a funny number have the form $3^{k} \cdot p$ for $k<5$?