Question

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

• A. 28
• B. 21
• C. 15
• D. 14

Answer 1

The correct option is A

28

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:

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

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

Now, $x\times (x+16)=x^2+16x$ and $(x+2)(x+14)=x^2+16x+28$

Hence $(x+2)(x+14)-x\times (x+16)=x^2+16x+28-(x^2+16x)=x^2+16x+28-x^2-16x=28$