a+b+c+d+e+f+g+h+i= 1+2+.....+9 = 10*9/2 =45

a+b+c = d+e+f= g+h+i = 45/3 = 15

each column, each row, and each of the two diagonals add up to 15

Also, (a+e+i)+(c+e+g)+(b+e+h)+(d+e+f) = 4*15

(a+b+c+d+e+f+g+h+i)+3e = 60

45+3e=60
e=5

Given a=6 and c= 2
b= 15-6-2=7

b+e+h= 15
7+5+h=15
h=3

We have 5 odds and 4 evens

If top middle is even, the sum is even but the sum of either middle or bottom rows will end up being odd. This means the top middle should be odd but this can't be 1 or 3 since then the sum will be too small for the other digits to work with 9

If top middle is 5, the sum is 13 but again 9 cannot be under 6, 5 or 2 (if it is under 2 we will need another 2 to get sum 13)

So the top middle can be 7 (sum 15). The next two rows can be 9,5,1 and 4,3,8

A little manipulation will give

6 7 2
1 5 9
8 3 4

Answer is B

Posted from my mobile device
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The average value in one cell of the table is 5. So on average, each row sums to 15. But each row-sum is the same. They can't all be above average or below-average (it's never possible for everything in a set to be above or below the average of the set), so each row must sum to exactly 15, and the same is true of each column and diagonal.

So the top row is 6 7 2. Now we can think about the middle column: it has a '7' in it, so the other two entries sum to 8. They can't be 4 and 4, 2 and 6, or 1 and 7, because we can't repeat numbers. They must be 5 and 3, in some order. And if the middle entry were 3, we get an impossible situation looking at either diagonal (we have 6 and 3, and to get 15 as a sum, we'd need another 6, or looking at the other diagonal, we have 2 and 3, which means we need to put a 10 in the bottom left, which we can't do).

So we must have a '5' in the middle, and a 3 in the bottom middle cell.

You can instead notice right away that '5' must be in the middle no matter what. Nick demonstrated that algebraically, but you can see why that must be true without algebra. Say we put '4' in the middle. There needs to be a '1' somewhere -- in a row, column, or diagonal. That row, column or diagonal needs to sum to 15, so the missing value needs to be 10, and we can't put '10' in this grid. So we can't have '4' in the middle, nor anything smaller, and analogously, we can't have '6' or higher, because we need to put a '9' somewhere, and then we get a sum that is too large. Once we know '5' is in the middle, filling in the top row then lets us deduce the middle-bottom value almost immediately.
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