Mystery number...

Note that the sum of these five numbers is five times the middle number (i.e. Brenda's number) so the sum can be expressed as $5r$ . Similayly, $q+r+s = 3r$ as Brenda's number is the middle of the 3 as well.

We have $5r = m^3$ and $3r = n^2$ where $m$ and $n$ are positive integers. Evidently, $5r$ must be divisible by $5^{3a}$ to be a perfect cube where $a$ is a positive integer. This means $r$ is divisible by $5^{3a-1}$ such that $3a-1>0$ . Similarly, $3r$ must also be divisible by $3^{2b}$ to be a perfect square where $b$ is a positive integer. So $r$ is divisible by $3^{2b-1}$ where $2b-1>0$ .

It follows that $r$ must have 3 and 5 as prime factors. Suppose $r = 5^{3a-1} \times 3^{2b-1}$ . This means $5r = m^3 = 5^{3a} \times 3^{2b-1}$ so $(2b-1)$ must be divisible by 3. Also $3r =n^2= 5^{3a-1} \times 3^{2n}$ so $(3a-1)$ must be divisible by 2. We find that an acceptable solution is $(a, b) = (1, 2)$ .

Now $r = 5^{3-1} \times 3^{2 \times 2 - 1} = 5^2 \times 3^3 = 675$ . We subconsciously made an assumption earlier that $r$ has 3 and 5 as its ONLY prime factors and it is a valid assumption as if $r=675$ has any other prime factors, even if it is 2, $r$ would exceed 1000 which is against the question. So the sum of the five numbers is $675 \times 5 = \boxed{3375}$ .