Make Every Row, Column & Diagonal Add Up To The Same Value!

If you are stuck, scroll down to the next section where I break down exactly how we can logically solve this problem…

How Can We Solve This?

Although this question may seem difficult, there is actually a logical way behind how we can solve this problem fast and effectively. First, we know that every row, column and diagonal has to add up to a certain value. Note: For my explanation, I will refer to boxes using the coordinate system where +5 is (1,1) and +3 is (3,1), etc. Right now, we are given 3 values of the diagonal starting at (1,4) or -7. Furthermore, the current sum of the diagonal is -6. More importantly, if we look at box (4,1), we can make a claim that whatever value goes into (4,1) will be incorporated in 3 sums: the bottom row with +5, the right column with +1 and the diagonal. Currently, the diagonal has -6. If we look at the bottom row, we need to make its current sum -6 as well so that the value in (4,1) can ensure that the sums are the same. Meaning, in the bottom row, we need to add -5 to make the sum 0 and then -6 to add the same negativity as the diagonal. To put this into more of a logical perspective, what’s really happening is that our diagonal’s sum can be represented as -6 + x. However, the sum for the bottom row is currently 5 + x. Thus, we have to make sure the equations become the same. So, to do so, we split -11 between coordinates (2,1) and (3,1). This way, both of our equations become -6 + x, which means no matter what the value of x is, the sum will be the same for the diagonal and the bottom row. Now, if you already tried to solve this problem using guess and check, you would have realized that -5 and -6 always go in the bottom row, and this is the logic behind why it happens. However, what we don’t know is the position of the values of -5 and -6 as they could go into (2,1) and (3,1) or (3,1) and (2,1) respectively. Now, let’s look at our findings:

Our Current Findings…

Moving on, let’s assume that box (2,1) is -5 and box (3,1) is -6. Furthermore, let’s assume a value of x. Since we incorporated a lot of negatives and we already have a sum of -6, the best thing to do would be to add a positive value. Otherwise, we wouldn’t have enough negatives to distribute throughout the rest of the square.

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