How can you place the numbers 1 through 9 in a 3x3 grid such that every row, column, and the two diagonals
all add up to 15?
To solve this riddle, it first seems logical to put the 5 in the middle square becuase it is the median and mean of the numbers from 1 to 9 (and also the average of any 3 numbers adding up to 15).
The next thing to do is place the 1 since it's the smallest and will thus be likely to quickly constrain what we can do afterward. We can try to place the 1 in either a side square or a corner square, but once we place it, it forces us to place the 9 on the opposite side. At this point, we just play around with the remaining numbers to see if we can work out a solution (there aren't many possibilities remaining for where we can place the other numbers).
Going along these lines, we'll quickly get to a solution.
This image shows how to place the numbers.
Consider the following explanation for why 1=2 :
|1. Start out||Let y = x|
|2. Multiply through by x||xy = x2|
|3. Subtract y2 from each side||xy - y2 = x2 - y2|
|4. Factor each side||y(x-y) = (x+y)(x-y)|
|5. Divide both sides by (x-y)||y = x+y|
|6. Divide both sides by y||y/y = x/y + y/y|
|7. And so...||1 = x/y + 1|
|8. Since x=y, x/y = 1||1 = 1 + 1|
|8. And so...||1 = 2|
How is this possible?
Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0.
This is an invalid mathematical operation (division by 0), and so by not followinng basic mathematical
rules, we are able to get strange results like these.