Explanation

This question resembles a popular puzzle. The substance here is non-algebraic here and naturally forks or prods us into analysis by cases. Indeed, any puzzle of this nature is suited to analysis by cases, because one unique case will prevail, while every other attractive-seeming case will ultimately run into a disqualifying contradiction. Let's turn to the data statements, separately first.

Statement (1) tells us that the upper-left and bottom-center boxes of the table are each 11. Because of the non-repetition rule of 11's, that actually means that , the square we are looking for, is the only other square that could be 11, it's the only square outside a row or column of the upper-left and bottom-center squares. So it could be 11. That's Case I. But we can form a Case II in which it's not 11, because the digit 5 may occur twice in a row or column, and that's a lot of potential extra 5's. E.g.,



So we have at least two possible values of , and hence we cannot answer the question definitively. Statement (1) is insufficient.

Statement (2) tells us that a jumble of numbers multiplied together gives us 375. All the numbers at hand are prime, so let's do a prime factorization of 375. The general principle is that when you have multiplication or division involving an integer with a large-ish absolute value, you should consider using prime factorization. The number is divisible by 375, so long division gives us



So this group, definitively, consists of one 3 and three 5's. So we are talking about the following shaded squares:



For most people, it would be wise to sketch this on the note board, since the notation of the variables and subscripts on the screen is prone to confusion. Our objective now, as is our objective generally when analyzing by cases in Data Sufficiency, is to obtain cases that are both allowed but yield different answers to the question, indicating insufficiency. Our square of interest cannot be a 5, because, wherever we try to put the 5's in the shaded squares, it would lead to three 5's in a row or column with the shaded squares. It could easily be 11, since none of the numbers in its column or row are 11's. So that's one case. The remaining question is whether it could be 3, forming another case. We don't know what shaded square the 3 is in, but regardless, putting a 3 in our square of interest would cause an illegal double in the row or column along which the already existing 3 is. Therefore, there is only one possible case, the square of interest must be an 11. We can answer the question definitively, so we have sufficient information.

The correct answer is (B).