Final Round

Suppose $n$ is divisor primary. Then $n$ cannot have an odd divisor $d\ge 5$. Indeed, for such a divisor, both $d-1$ and $d+1$ are even. Because $d-1>2$, these are both composite numbers and that would contradict the fact that $n$ is divisor primary. The odd divisors $1$ and $3$ can occur, because the integer $3$ itself is divisor primary.

Because of the unique factorisation in primes, the integer $n$ can now only have some factors $2$ and at most one factor $3$. The number $2^6=64$ and all its multiples are not divisor primary, because both $63=7\cdot 9$ and $65=5\cdot 13$ are not prime. Hence, a divisor primary number has at most five factors $2$. Therefore, the largest possible number that could still be divisor primary is $3\cdot 2^5=96$.

We now check that $96$ is indeed divisor primary: its divisors are $1$, $2$, $3$, $4$, $6$, $8$, $12$, $16$, $24$, $32$, $48$, and $96$, and these numbers are next to $2$, $3$, $2$, $3$, $5$, $7$, $11$, $17$, $23$, $31$, $47$, and $97$, which are all prime. Therefore, the largest divisor primary number is $96$. $\square$