Pizza Hut Math (equal-sides sets of numbers with same sum and product) Pizza Hut, in honor of $\pi$ day, posted this problem:

Our school’s puzzle-club meets in one of the schoolrooms every Friday after school.
Last Friday, one of the members said, “I’ve hidden a list of numbers in this envelope that add up to the number of this room.” A girl said, “That’s obviously not enough information to determine the number of the room. If you told us the number of numbers in the envelope and their product, would that be enough to work them all out?”
He (after scribbling for some time): “No.” She (after scribbling for some more time): “well, at least I’ve worked out their product.”
What is the number of the school room we meet in?”

Is there a non-brute-force way of doing this problem? Additionally, I've found multiple same-sized sets of numbers with the same product and sum, so I imagine I am utterly misunderstanding the problem anyway (e.g. 2 triples that sum to 23 and multiply to 360, 2 triples that sum to 22 and multiply to 360, 2 triples that sum to 21 and multiply to 240, etc.). 

Well, here is my solution (tl;dr my answer is 12)
First, we are looking for a number. We can write all possible unique "sum representations" of that number with different length. Every representation is a list of sorted numbers. For example 3 can be represented as [[3], [1 2], [1 1 1]]. Next, for every representation we can calculate product of the elements in it, for 3 it will be [3, 2, 1]. Also for every representation we can find number of elements in it: [1, 2, 3]. We in the "sum representations" of a number we want to find such representations that have same product of elements, but only one. I am sorry, I don't know how to explain it better (English is not my native). But let me explain why 23 is not an answer. 
Indeed, [4 9 10] and [5 6 12] are two sum-representations of 23 with product equals to 360. But, there is another representations of 23: [1 2 8 12] and [2 2 3 16] that give product 192. Thus, girl would not be able to guess the product in case of room number 23, because there is a lot of groups of representation (226) that will give the same product. 
Long story short, I was not able to find a number lower then 12 that would give me two representations with the same product. 12=1+3+4+4=2+2+2+6. 13 already have two different repeated products: 36 ([1 6 6] and [2 2 9]) and 48 ([1 1 3 4 4] and [1 2 2 2 6]). As you may notice, any other number bigger than 13 will have at least two representations with the product 36 and at least two with the product 48. So, 12 is the only solution. 
I am sorry for quite complicated explanation. I am not doing math on professional level, so I do not know all terminology of the number theory. I will be glad to answer any questions. 
