<p>Ok, so first off, I think the answers are B. 10, 5940, and 51. For the first one, if you just multiply (xy)(yz)(xz) you obtain the answer for (x^2)(y^2)(z^2) = (2)(5)(10) = 100, but (x^2)(y^2)(z^2) is really equal to (xyz)^2 so then you take the square root of 100 to get an answer of 10.
Next for the second question, I used a trial and error method. I assumed that W, X, Y, and Z must between 0 - 9 because those are all single digits. Now you know that W is 5 greater than Z and since W also 1 greater than Y, Y must be 5 greater than Z as well. But if you just look at that, W and Y, based on the amount they must be greater than Z is already 9 and since I assumed that all those numbers must be between 0-9, I realized that Z=0 so that X, the sum of X Y and Z must be between 0 -9 so therefore I concluded that W = 5, X = 9, Y = 4, and Z = 0.
Finally, for the last question, I immediately noticed that the last 2 digits on the total sum was equal to the 2 digits in 23, so I figured that p must be 1. Then I subtracted 23 from 4523 to get that 90n = 4500 and so i divided and found that n = 50. Now 50 + 1 = 51 so therefore n + p = 51 for one possible solution</p>