Calendar Magic (2) - STD 8 - Chap: 3
📅 Calendar Magic: 3×3 Square | Diagonal Corner Products Difference = 28
Click any date to center a 3×3 block | Thick border around the 3x3 box | Four corners are highlighted in yellow
📆 Select Month & Year
🔢 Four Corner Numbers (Highlighted in Yellow)
Top-Left
-
Top-Right
-
Bottom-Left
-
Bottom-Right
-
Diagonal 1 (TL × BR): -
Diagonal 2 (TR × BL): -
Difference (|D1 - D2|) = -
Click "Next Step" to begin the proof
💡 Algebraic Proof:
\[ \text{Let the center number be } n \] \[ \text{In a calendar: right } +1,\ \text{down } +7 \] \[ \begin{aligned} \text{Top-Left} &= n - 8 \\ \text{Top-Right} &= n - 6 \\ \text{Bottom-Left} &= n + 6 \\ \text{Bottom-Right} &= n + 8 \end{aligned} \] \[ \begin{aligned} \text{Diagonal 1 (TL × BR)} &= (n-8)(n+8) = n^2 - 64 \\ \text{Diagonal 2 (TR × BL)} &= (n-6)(n+6) = n^2 - 36 \end{aligned} \] \[ \text{Difference} = (n^2 - 64) - (n^2 - 36) = -28 \] \[ \text{Absolute difference} = | -28 | = 28 \]
✅ Therefore, for any 3×3 block, the absolute difference of diagonal corner products is 28!
\[ \text{Let the center number be } n \] \[ \text{In a calendar: right } +1,\ \text{down } +7 \] \[ \begin{aligned} \text{Top-Left} &= n - 8 \\ \text{Top-Right} &= n - 6 \\ \text{Bottom-Left} &= n + 6 \\ \text{Bottom-Right} &= n + 8 \end{aligned} \] \[ \begin{aligned} \text{Diagonal 1 (TL × BR)} &= (n-8)(n+8) = n^2 - 64 \\ \text{Diagonal 2 (TR × BL)} &= (n-6)(n+6) = n^2 - 36 \end{aligned} \] \[ \text{Difference} = (n^2 - 64) - (n^2 - 36) = -28 \] \[ \text{Absolute difference} = | -28 | = 28 \]
✅ Therefore, for any 3×3 block, the absolute difference of diagonal corner products is 28!