Divisors

We mean to find all positive integers $k$ , whose greatest divisor, besides itself, is $91$ .

We know that $91 = 13 * 7$ . The integers we want to find are multiples of $91$ , so $k = 7 * 13 * n$ . Since $91$ is the greatest divisor, $n \leq 7$ , because if $n > 7$ , then the greatest divisor of $k$ is $13n$ .

We can also conclude that $n$ has to be a prime number, otherwise, we would have $n = a * b$ ( $a > 1$ and $b > 1$ ) and both $b * 13 * 7$ and $a * 13 *7$ would be greater than $13 * 7$ and divisors of $k$ .

As such, the possible values of $n$ are $2, 3, 5, 7$ which would correspond to the integers $182, 273, 455, 637$ . So the answer is $\boxed4$ .

Let $N$ be a positive integer, $g$ be the greatest divisor of $N~(g\neq N)$ and $s$ be the smallest divisor of $N~(s\neq 1)$ .

$N=s \times g$

In order to be the smallest divisor, $s$ must be a prime and smaller than or equal to the smallest divisor of $g$ .

In this case, $g=91=7 \times 13$ . Therefore, $s \leq 7$ and $s$ is prime. The solutions for $s$ are 2, 3, 5, 7, so there are $\boxed{4}$ positive integers such that their greatest divisor, different from itself, is $91$ .

How is this level 5, by the way? 91 can be factored into 7 x 13. The only other integers that have 7 13 as the largest divisor apart from itself must be 7 13 x. This means that the x cannot be factored anymore (7 13 6's largest divisor would be 7 13*3), and cannot be larger than 7. This leaves only the primes 2, 3, 5, 7.