
1. A positive integer $n$ is called divisor primary if for every positive divisor $d$ of $n$ at least one of the numbers $d-1$ and $d+1$ is prime. For example, 8 is divisor primary, because its positive divisors $1,2,4$, and 8 each differ by 1 from a prime number $(2,3,5$, and 7 , respectively), while 9 is not divisor primary, because the divisor 9 does not differ by 1 from a prime number (both 8 and 10 are composite).

(a) Which odd numbers can occur as the divisor of a divisor primary number?

(b) Determine the largest divisor primary number.

## 1. Version for klas 5 \& klas 4 and below

(a) Suppose $n$ is divisor primary. Then $n$ cannot have an odd divisor $d \geqslant 5$. 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.

(b) 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 .

