Funny Numbers!

We have several observations:

• If $n$ is even, then it is not funny, since 2 is a divisor of $n$ but $2+2 = 4$ is not prime.
• If $n$ has a prime divisor in the form $3k+1$ , then it is not funny, since $(3k+1) + 2 = 3k+3$ is not prime.
• If $n$ has two prime divisors (counting multiplicity) in the form $3p+2, 3q+2$ , then it is not funny, since $(3p+2)(3q+2)$ divides $n$ and $(3p+2)(3q+2) + 2 = 9pq + 6p + 6q + 6$ is a multiple of 3, hence not prime.

Thus either $n = 3^a$ or $n = 3^a (3p+2)$ , where $a$ is a nonnegative number and $3p+2$ is prime. (If not, then $n$ has more than one prime divisors that aren't 3; either some of them are $3k+1$ , or all of them are $3k+2$ and hence there are at least two of them.)

Observe that $n = 135 = 3^3 \cdot 5$ satisfies this condition, with 8 positive divisors. Also observe that $3^5 + 2 = 245$ is not prime (multiple of 5), so $a \le 4$ . Thus $n = 3^a$ is impossible since it only has $a+1 < 8$ positive divisors.

Now, $n = 3^a (3p+2)$ has $2(a+1)$ positive divisors, so we need $a \ge 3$ . Thus $3p+2, 3(3p+2), 9(3p+2), 27(3p+2)$ are all divisors of $n$ . Note that they are congruent to $3p+2, 2(3p+2), 3(3p+2), 4(3p+2)$ modulo 5. If $3p+2 \neq 5$ , then $3p+2$ is not divisible by 5, and thus these four form an almost complete congruence classes modulo 5 (they are 1, 2, 3, 4 modulo 5 in some order). Suppose $3^b(3p+2) \equiv 3 \pmod 5$ . Then $3^b(3p+2) + 2 \equiv 0 \pmod 5$ , and clearly since $3p+2 \ge 5$ , $3^b(3p+2) + 2 \neq 5$ . Thus it is divisible by 5 and hence not prime.

Thus the assumption $3p+2 \neq 5$ is false, so $3p+2 = 5$ . It can be verified that $3^4 \cdot 5 + 2 = 407$ is not prime (multiple of 11), so the $n$ with the most positive divisors is $3^3 \cdot 5 = 135$ .