First of all, if the product of the ages was 2450, then I needed to apply prime factorization to that number.
This got me 2*5*5*7*7. The 3 ages must be products of these numbers.
Multiplying them, and then dividing the product by 2 to get the assistant's age, I got:
Ages + assistant
2, 25, 49 + 38
10, 5, 49 + 32
14, 5, 35 + 27
10, 7, 35 + 26
14, 7, 25 + 23
Now, none of these helped, and I remembered from the last problem like this that there needed to be a combination that came out with an assistant age exactly like one of these, so I started multplying 3 factors together, and got:
50, 7, 7 + 32
This would explain why the assistant needed more info, since there are now two possible combinations with his age; 32.
Therefore, when the professor said she was older than the 3 people she dined with, we look at the combinations again:
10, 5, 49 + 32
50, 7, 7 + 32
As soon as she said she was older, the answer became obvious to the assistant. One set has a high number of 49, and the other set has 50. Only one set can be correct, and since the apprentice knows her age, he can ignore the set that has her age (making one of the diners the same age.) It has to be 10, 5, 49, and since the difference between the highest numbers is 1 - which doesn't provide enough range for ages higher than the professor's - then the professor is 50.
Make sense?