Take nine numbers forming a square in a calendar and mark the four numbers at the corners. Then the difference between the squares of the diagonal pairs is always ____.

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Solution

The correct option is A
56

Let the number at the top left corner of the square of nine numbers in a calendar be 'x'.

Then such a square in the calendar would be of the form:

$\begin{matrix} x & x + 1 & x + 2 x + 7 & x + 8 & x + 9 x + 14 & x + 15 & x + 16 \end{matrix}$

The four numbers at the corners are $x, x + 2, x + 14, and x + 16$ , in which $x and x + 16$ form one diagonal pair and $x + 2 and x + 14$ form another diagonal pair.

Now, $x^{2} + (x + 16)^{2} = x^{2} + (x^{2} + 32 x + 256) = x^{2} + x^{2} + 32 x + 256 = 2 x^{2} + 32 x + 256$

and $(x + 14)^{2} + (x + 2)^{2} = (x^{2} + 28 x + 196) + (x^{2} + 4 x + 4) = x^{2} + 28 x + 196 + x^{2} + 4 x + 4 = 2 x^{2} + 32 x + 200$

So, $(x^{2} + (x + 16)^{2}) - ((x + 14)^{2} + (x + 2)^{2}) = (2 x^{2} + 32 x + 256) - (2 x^{2} + 32 x + 200) = 2 x^{2} + 32 x + 256 - 2 x^{2} - 32 x - 200 = 56$

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