Ugh. I just had this whole thing typed in and lost it. Ah well.
I have a systematic way of solving the problem, but I admit that it's long and trial and error is much shorter. Here it is, briefly. You can easily fill in the steps.
We want to find a z such that z = (x^a)(y^b) where x and y are distinct primes, a and b are positive integers such that a + b = 6, and ax + by = 28.
why . shouldn't it be . also why is the expression for the sum of divisors ax+by. i guess it should be product of two geometric series?? .
We can say that b = 6 - a, so the second condition becomes ax + (6 - a)y = 28. We need to solve this equation.
First note that, according to the problem statement x and y are interchangeable. Let's start then, by letting x = 2. When we do this we run into a contradiction. So x is not 2 and since x and y are interchangeable, neither is y.
So x and y are both (distinct) odd primes.
Now, since ax + (6 - a)y = 28 either ax and (6 - a)y are both even or they are both odd.
Assume ax and (6 - a)y are both even. Then a must be even since x and y are odd. Let a = 2n. Then n = 1 or n = 2. Both cases create a contradiction, so ax and (6 - a)y both cannot be even.
So ax and (6 - a)y must both be odd. Thus a is odd. Let a = 2n - 1. Then n = 1, 2, or 3. Since x and y are interchangeable n = 1 and n = 2 both provide the same result. n = 3 gives a contradiction. Thus let n = 1. So we (finally) have that x + 5y = 28.
Now, x and y are distinct primes, not equal to 2. Thus x = 3, y = 5; x = 5, y = 7; x = 3, y = 7, etc. Of these only x = 3, y = 5 can be a solution since all other possibilities give x + 5y > 28.
Thus the only solution is x = 3, y = 5 and since n = 1 we know that a = 1 and b = 5, which gives z = (3^1)(5^5) = 9375.
One would hope for a shorter solution method....
-Dan